MECHANICS 


OF 


MATERIALS 


BY 

MANSFIELD     MERRIMAN 

MEMBER  OF  AMERICAN  SOCIETY  OF  CIVIL  ENGINEERS 


ELEVENTH  EDITION 
TOTAL  ISSUE,  FIFTY-EIGHT  THOUSAND 


NEW   YORK 

JOHN  WILEY  &  SONS,  INC. 
LONDON:  CHAPMAN  &  HALL,  LIMITED 


/V\b- 


COPYRIGHT,  1885,  1890,  1895,  1905 

BY 
MANSFIELD   MERRIMAN 

First  Edition,  June,  1885 
Second  Edition,  revised  and  enlarged,  April,  1890 

Third  Edition,  February,  1891 

Fourth  Edition,  June,  1892;  reprinted,  1893 

Fifth  Edition,  February,  1894;   reprinted,  1895 

Sixth  Edition,  enlarged,  September,  1895;   reprinted,  1896 

Seventh  Edition,  July,  1897 
Eighth  Edition,  February,  1899 

Ninth  Edition,  June,  1900;   reprinted,  1901,  1902,  1903  (twice),  1904,  190$ 
Tenth  Edition,  rewritten  and  enlarged,  September,  1905; 

reprinted,  1906,  1907,  1908,  1909,  1910,  1911,  1912 

Eleventh  Edition,  March,  1914;    reprinted,  1915,  1916 

ALL  RIGHTS  RESERVED 


4 


PRESS  OF 
r /2I  BRAUNWORTH   &  CO. 


BOOK  MANUFACTURERS 
BROOKLYN.    N.  Y. 


PREFACE  TO  TENTH  EDITION 

SINCE  1885,  when  the  first  edition  of  this  work  was  published, 
many  advances  have  been  made  in  the  subject  of  Mechanics  of 
Materials.  Some  of  these  have  been  noted  in  the  additions 
to  subsequent  editions,  but  to  record  and  correlate  them  properly 
it  has  now  become  necessary  to  rewrite  and  reset  the  book.  In 
doing  so  the  author  has  endeavored  to  keep  the  facts  of  experi- 
ment and  practice  constantly  in  view,  for  the  theory  of  the  subject 
is  merely  the  formal  expression  and  generalization  of  observed 
phenomena.  The  subject,  indeed,  no  longer  consists  of  a  series 
of  academic  exercises  in  algebra  and  rational  mechanics,  but 
it  is  indispensably  necessary  that  the  phenomena  of  stress  should 
be  clearly  understood  by  the  student.  While  laboratory  work 
is  a  valuable  aid  to  this  end,  it  is  important,  in  the  opinion  of 
the  author,  that  no  recitation  or  lecture  should  be  held  without 
having  test  specimens  at  hand  with  which  to  illustrate  the  phys- 
ical phenomena. 

The  same  general  plan  of  treatment  has  been  followed  as 
before,  but  the  subdivisions  are  somewhat  different,  and  the  fifteen 
chapters  of  the  last  edition  have  been  increased  to  nineteen. 
The  statement  of  average  values  of  the  principal  materials  of 
engineering  has  proved  so  advantageous  to  students  that  it  is 
here  also  followed.  Numerous  numerical  examples  are  given 
in  the  text  to  exemplify  the  formulas  and  methods,  these  generally 
relating  to  cases  that  arise  in  practice.  To  encourage  students 
to  think  for  themselves,  one  or  more  problems  are  given  at  the 
end  of  each  article;  for  the  experience  of  the  author  has  indicated 
that  the  solution  of  many  numerical  exercises  is  required  in 
order  that  students  may  become  well  grounded  in  theory. 

Most  of  the  topics  of  the  last  edition  have  been  treated  in  a 
fuller  manner  than  before.  The  subjects  of  impact  on  bars 
and  beams,  resilience  and  work,  and  apparent  and  true  stresses 

456298  Ui 


iv  PREFACE 

have  been  much  changed  with  the  intention  of  rendering  the 
presentation  more  clear  and  accurate.  Among  many  new  topics 
introduced  are  those  of  economic  sections  for  beams,  moving 
loads  on  beams,  constrained  beams  with  supports  on  different 
levels,  the  torsion  of  rectangular  bars,  compound  columns  and 
beams,  reinforced-concrete  beams,  plates  under  concentrated 
loads,  internal  friction,  rules  for  testing  materials,  and  elastic- 
electric  analogies.  A  few  changes  in  algebraic  notation  have 
been  made  in  order  that  similar  quantities  may  always  be  desig- 
nated by  letters  of  the  same  type;  Greek  letters  are  used  only 
for  angles  and  abstract  numbers. 

Compared  with  the  ninth  edition,  the  number  of  articles  has 
been  increased  from  151  to  188,'  the  number  of  tables  from  8 
to  20,  the  number  of  cuts  from  85  to  250,  and  the  number  of 
problems  from  222  to  305.  Although  the  length  of  each  page 
has  been  increased  eight  percent  and  smaller  type  has  been  used 
for  formulas  and  problems,  the  number  of  pages  has  been  in- 
creased from  378  to  518.  While  the  main  purpose  in  rewriting 
and  enlarging  the  book  has  been  to  keep  it  abreast  with  modern 
progress,  the  attempt  has  also  been  made  to  present  the  subject 
more  clearly  and  logically  than  before,  in  order  both  to  advance 
the  interests  of  sound  engineering  education  and  to  promote  sound 
engineering  practice. 

NOTE 

In  this  eleventh  edition  the  subject  of  testing  of  materials  is 
treated  in  the  new  Chapter  XX.  New  matter  on  influence  lines, 
curved  beams,  springs,  and  moment  of  inertia  is  given  in  Chapter 
XVIII.  A  few  minor  changes  have  been  made  in  other  pages, 
and  all  known  typographic  errors  have  been  corrected. 

M.  M. 


In  this  impression  of  the  eleventh  edition  Arts.  86  and  160 
are  rewritten,  and  a  few  minor  changes  are  made  here  and  there. 

M.  M. 


CONTENTS 

CHAPTER  I 
ELASTIC  AND  ULTIMATE  STRENGTH 

PAGE 

ART.    1.    SIMPLE  AXIAL  STRESSES i 

2.  THE  ELASTIC  LIMIT 4 

3.  ULTIMATE  STRENGTH. 6 

4.  TENSION 9 

5.  COMPRESSION n 

6.  SHEAR 14 

7.  WORKING  UNIT-STRESSES 16 

8.  COMPUTATIONS  AND  EQUATIONS 19 

CHAPTER  II 
ELASTIC  AND   ULTIMATE  DEFORMATION 

ART.    9.    MODULUS  OF  ELASTICITY 23 

10.  ELASTIC  CHANGE  OF  LENGTH 25 

11.  ELASTIC  LIMIT  AND  YIELD  POINT 27 

12.  ULTIMATE  DEFORMATIONS 30 

13.  CHANGES  IN  SECTION  AND  VOLUME 32 

14.  WORK  IN  PRODUCING  DEFORMATION  . 35 

15.  SHEARING  MODULUS  OF  ELASTICITY 37 

16.  HISTORICAL  NOTES 39 

CHAPTER  III 
MATERIALS  OF  ENGINEERING 

ART.  17.    AVERAGE  WEIGHTS 42 

18.  PLASTICITY  AND  BRITTLENESS 43 

19.  TIMBER 46 

20.  BRICK 48 

21.  STONE 50 

22.  MORTAR  AND  CONCRETE 52 

23.  CAST  IRON 55 

24.  WROUGHT  IRON 57 

25.  STEEL 60 

26.  OTHER  MATERIALS 66 

v 


vi  CONTENTS 

CHAPTER  IV 

CASES   OF  SIMPLE  STRESS 

PAGE 

ART.  27.  STRESS  UNDER  OWN  WEIGHT 69 

28.  BAR  OF  UNIFORM  STRENGTH 71 

29.  ECCENTRIC  LOADS 72 

30.  WATER  AND  STEAM  PIPES 75 

31.  THIN  CYLINDERS  AND  SPHERES 77 

32.  SHRINKAGE  OF  HOOPS 79 

33.  INVESTIGATION  OF  RIVETED  JOINTS 80 

34.  DESIGN  OF  RIVETED  JOINTS 83 

CHAPTER  V 
GENERAL  THEORY   OF  BEAMS 

ART.  35.     DEFINITIONS 87 

36.  REACTIONS  OF  SUPPORTS 88 

37.  THE  VERTICAL  SHEAR 90 

38.  THE  BENDING  MOMENT 93 

39.  INTERNAL  STRESSES  AND  EXTERNAL  FORCES 96 

40.  NEUTRAL  SURFACE  AND  Axis 98 

41.  SHEAR  AND  FLEXURE  FORMULAS 101 

42.  CENTER  OF  GRAVITY 103 

43.  MOMENTS  OF  INERTIA 105 

44.  ROLLED  BEAMS  AND  SHAPES 108 

45.  ELASTIC  DEFLECTIONS 112 


CHAPTER  VI 

SIMPLE  AND   CANTILEVER  BEAMS 

A.RT.  46.  SHEAR  AND  MOMENT  DIAGRAMS 116 

47.  MAXIMUM  SHEARS  AND  MOMENTS 119 

48.  INVESTIGATION  OF  BEAMS   .     .     . 121 

49.  SAFE  LOADS  FOR  BEAMS 124 

50.  DESIGNING  OF  BEAMS 125 

51.  ECONOMIC  SECTIONS 127 

52.  RUPTURE  OF  BEAMS 130 

53.  MOVING  LOADS 132 

54.  DEFLECTION  OF  CANTILEVER  BEAMS 135 

55.  DEFLECTION  OF  SIMPLE  BEAMS 138 

56.  COMPARATIVE  STRENGTH  AND  STIFFNESS 141 

57.  CANTILEVER  BEAMS  OF  UNIFORM  STRENGTH 143 

58.  SIMPLE  BEAMS  OF  UNIFORM  STRENGTH 146 


CONTENTS  vii 

CHAPTER  VII 
OVERHANGING  AND  FIXED  BEAMS 

PAGB 

ART.  59.  BEAM  OVERHANGING  ONE  SUPPORT 149 

60.  BEAM  FIXED  AT  ONE  END 152 

61.  BEAM  OVERHANGING  BOTH  SUPPORTS 154 

62.  BEAM  FIXED  AT  BOTH  ENDS 156 

63.  COMPARISON  OF  BEAMS 158 

64.  SUPPORTS  ON  DIFFERENT  LEVELS 159 

65.  CANTILEVER  WITH  CONSTRAINT 163 

66.  SPECIAL  DISCUSSIONS 164 


CHAPTER  VIII 

CONTINUOUS  BEAMS 

ART.  67.     GENERAL  PRINCIPLES 168 

68.  METHOD  OF  DISCUSSION 171 

69.  THEOREM  OF  THREE  MOMENTS 173 

70.  EQUAL  SPANS  WITH  UNIFORM  LOAD 175 

71.  UNEQUAL  SPANS  AND  LOADS 177 

72.  SPANS  WITH  FIXED  ENDS 179 

73.  CONCENTRATED  LOADS  . 180 

74.  SUPPORTS  ON  DIFFERENT  LEVELS 182 

75.  THE  THEORY  OF  FLEXURE 184 


CHAPTER  IX 
COLUMNS   OR  STRUTS 

ART.  76.  CROSS-SECTIONS  OF  COLUMNS 188 

77.  DEFINITIONS  AND  PRINCIPLES 190 

78.  EULER'S  FORMULA  FOR  LONG  COLUMNS 192 

79.  EXPERIMENTS  ON  COLUMNS 196 

80.  RANKINE'S  FORMULA 200 

81.  INVESTIGATION  OF  COLUMNS 203 

82.  SAFE  LOADS  FOR  COLUMNS ...  205 

83.  DESIGNING  OF  COLUMNS 206 

84.  THE  STRAIGHT-LINE  FORMULA 208 

85.  OTHER  COLUMN  FORMULAS      ...  211 

86.  ECCENTRIC  LOADS  ON  PRISMS        .     .  214 

87.  ECCENTRIC  LOADS  ON  COLUMNS 217 

88.  ON  THE  THEORY  OF  COLUMNS 220 


vi  CONTENTS 

CHAPTER  IV 

CASES   OF  SIMPLE  STRESS 

PAGE 

ART.  27.     STRESS  UNDER  OWN  WEIGHT 69 

28.  BAR  OF  UNIFORM  STRENGTH 71 

29.  ECCENTRIC  LOADS 72 

30.  WATER  AND  STEAM  PIPES 75 

31.  THIN  CYLINDERS  AND  SPHERES 77 

32.  SHRINKAGE  OF  HOOPS 79 

33.  INVESTIGATION  OF  RIVETED  JOINTS 80 

34.  DESIGN  OF  RIVETED  JOINTS 83 


CHAPTER  V 
GENERAL  THEORY   OF  BEAMS 

ART.  35.  DEFINITIONS 87 

36.  REACTIONS  OF  SUPPORTS 88 

37.  THE  VERTICAL  SHEAR 90 

38.  THE  BENDING  MOMENT 93 

39.  INTERNAL  STRESSES  AND  EXTERNAL  FORCES 96 

40.  NEUTRAL  SURFACE  AND  Axis 98 

41.  SHEAR  AND  FLEXURE  FORMULAS 101 

42.  CENTER  OF  GRAVITY 103 

43.  MOMENTS  OF  INERTIA 105 

44.  ROLLED  BEAMS  AND  SHAPES 108 

45.  ELASTIC  DEFLECTIONS 112 


CHAPTER  VI 
SIMPLE  AND   CANTILEVER  BEAMS 

46.  SHEAR  AND  MOMENT  DIAGRAMS 116 

47.  MAXIMUM  SHEARS  AND  MOMENTS 119 

48.  INVESTIGATION  OF  BEAMS 121 

49.  SAFE  LOADS  FOR  BEAMS 124 

50.  DESIGNING  OF  BEAMS 125 

51.  ECONOMIC  SECTIONS 127 

52.  RUPTURE  OF  BEAMS 130 

53.  MOVING  LOADS 132 

54.  DEFLECTION  OF  CANTILEVER  BEAMS 135 

55.  DEFLECTION  OF  SIMPLE  BEAMS 138 

56.  COMPARATIVE  STRENGTH  AND  STIFFNESS 141 

57.  CANTILEVER  BEAMS  OF  UNIFORM  STRENGTH 143 

58.  SIMPLE  BEAMS  OF  UNIFORM  STRENGTH 146 


CONTENTS  vii 

CHAPTER  VII 
OVERHANGING  AND  FIXED  BEAMS 

PAGE 

AKT.  59.  BEAM  OVERHANGING  ONE  SUPPORT 149 

60.  BEAM  FIXED  AT  ONE  END 152 

61.  BEAM  OVERHANGING  BOTH  SUPPORTS 154 

62.  BEAM  FIXED  AT  BOTH  ENDS 156 

63.  COMPARISON  or  BEAMS 158 

64.  SUPPORTS  ON  DIFFERENT  LEVELS 159 

65.  CANTILEVER  WITH  CONSTRAINT 163 

66.  SPECIAL  DISCUSSIONS 164 


CHAPTER  VIII 
CONTINUOUS  BEAMS 

A.RT.  67.  GENERAL  PRINCIPLES 168 

68.  METHOD  OF  DISCUSSION 171 

69.  THEOREM  or  THREE  MOMENTS 173 

70.  EQUAL  SPANS  WITH  UNIFORM  LOAD 175 

71.  UNEQUAL  SPANS  AND  LOADS 177 

72.  SPANS  WITH  FIXED  ENDS 179 

73.  CONCENTRATED  LOADS  .  180 

74.  SUPPORTS  ON  DIFFERENT  LEVELS 182 

75.  THE  THEORY  OF  FLEXURE 184 


CHAPTER  IX 

COLUMNS   OR   STRUTS 

ART.  76.  CROSS-SECTIONS  OF  COLUMNS 188 

77.  DEFINITIONS  AND  PRINCIPLES 190 

78.  EULER'S  FORMULA  FOR  LONG  COLUMNS 192 

79.  EXPERIMENTS  ON  COLUMNS 196 

80.  RANKINE'S  FORMULA 200 

81.  INVESTIGATION  OF  COLUMNS 203 

82.  SAFE  LOADS  FOR  COLUMNS ...  205 

83.  DESIGNING  OF  COLUMNS 206 

84.  THE  STRAIGHT-LINE  FORMULA 208 

85.  OTHER  COLUMN  FORMULAS 211 

86.  ECCENTRIC  LOADS  ON  PRISMS        .     .  214 

87.  ECCENTRIC  LOADS  ON  COLUMNS 217 

88.  ON  THE  THEORY  OF  COLUMNS 220 


viii  CONTENTS 

CHAPTER  X 

TORSION  OF  SHAFTS 

TJam 
ART.  89.    PHENOMENA  OF  TORSION 225 

90.  THE  TORSION  FORMULA 227 

91.  SHAFTS  FOR  TRANSMITTING  POWER 250 

92.  SOLID  AND  HOLLOW  SHAFTS 231 

93.  TWIST  OF  SHAFTS 233 

94      RUPTURE  OF  SHAFTS 235 

95.  STRENGTH  AND  STIFFNESS 237 

96.  SHAFT  COUPLINGS 239 

97.  A  SHAFT  WITH  CRANK 240 

98.  A  TRIPLE-CRANK  SHAFT 242 

99.  NON-CIRCULAR  SECTIONS 245 


CHAPTER  XI 
APPARENT  COMBINED  STRESSES 

ART.  100.    STRESSES  DUE  TO  TEMPERATURE 251 

101.  BEAMS  UNDER  AXIAL  FORCES 253 

102.  FLEXURE  AND  COMPRESSION 255 

103.  FLEXURE  AND  TENSION    . 259 

104.  ECCENTRIC  AXIAL  FORCES  ON  BEAMS 261 

105.  SHEAR  AND  AXIAL  STRESS 263 

106.  FLEXURE  AND  TORSION 266 

107.  COMPRESSION  AND  TORSION 268 

108.  HORIZONTAL  SHEAR  IN  BEAMS 269 

109.  LINES  OF  STRESS  IN  BEAMS 272 


CHAPTER  XII 
COMPOUND  COLUMNS  AND  BEAMS 

ART.  110.  BARS  OF  DIFFERENT  MATERIALS 276 

111.  COMPOUND  COLUMNS 279 

112.  FLITCHED  BEAMS 282 

113.  REINFORCED  CONCRETE  BEAMS 285 

114.  THEORY  OF  REINFORCED  CONCRETE  BEAMS 289 

115.  INVESTIGATION  OF  REINFORCED  CONCRETE  BEAMS   .     .     .     .292 

116.  DESIGN  OF  REINFORCED  CONCRETE  BEAMS 295 

117.  PLATE  GIRDERS 298 

118.  DEFLECTION  OF  COMPOUND  BEAMS 300 


CONTENTS  ix 

CHAPTER  XIII 
RESILIENCE  AND  WORK 

PASH 

A*T.  119.  EXTERNAL  WORK  AND  INTERNAL  ENERGY 303 

120.  RESILIENCE  OF  BARS 306 

121.  RESILIENCE  OF  BEAMS 308 

122.  RESILIENCE  IN  SHEAR  AND  TORSION 310 

123.  DEFLECTION  UNDER  ONE  LOAD 312 

124.  DEFLECTION  AT  ANY  POINT 314 

125.  DEFLECTION  DUE  TO  SHEAR 316 

126.  PRINCIPLE  OF  LEAST  WORK 330 


CHAPTER  XIV 
IMPACT  AND   FATIGUE 

ART.  127.  SUDDEN  LOADS  AND  STRESSES 324 

128.  AXIAL  IMPACT  ON  BARS 327 

129.  TRANSVERSE  IMPACT  ON  BEAMS 329 

130.  INERTIA  IN  AXIAL  IMPACT 331 

131.  INERTIA  IN  TRANSVERSE  IMPACT 334 

132.  VIBRATIONS  AFTER  IMPACT 338 

133.  EXPERIMENTS  ON  ELASTIC  IMPACT 341 

134.  PRESSURE  DURING  IMPACT 344 

135.  IMPACT  CAUSING  RUPTURE 346 

136.  STRESSES  DUE  TO  LIVE  LOADS 349 

137.  THE  FATIGUE  OF  MATERIALS 352 

138.  STRENGTH  UNDER  FATIGUE 355 


CHAPTER  XV 
TRUE  INTERNAL  STRESSES 

ART.  139.  PRINCIPLES  AND  LAWS 359 

140.  SHEAR  DUE  TO  NORMAL  STRESS 362 

141.  COMBINED  SHEAR  AND  AXIAL  STRESS 365 

142.  TRUE  STRESSES  FOR  BEAMS 367 

143.  NORMAL  STRESSES  DUE  TO  SHEAR 369 

144.  TRUE  STRESSES  IN  SHAFTS 371 

145.  PURE  INTERNAL  STRESS 373 

146.  INTERNAL  FRICTION 375 

147.  THEORY  OF  INTERNAL  FRICTION 378 


*  CONTENTS 

CHAPTER  XVI 
GUNS  AND  THICK  CYLINDERS 

PAOB 

A*T.  148.  PRINCIPLES  AND  METHODS ....  383 

149.  LAMP'S  FORMULAS 385 

150.  SOLID  GUNS  AND  THICK  PIPES 388 

151.  A  COMPOUND  CYLINDER 390 

152.  CLAVARINO'S  FORMULAS 392 

153.  BIRNIE'S  FORMULAS 394 

154.  HOOP  SHRINKAGE 39u 

155.  DESIGN  OF  HOOPED  GUNS 399 

CHAPTER  XVII 
ROLLERS,   PLATES,   SPHERES 

ART.  156.  CYLINDRICAL  ROLLERS 403 

157.  SPHERICAL  ROLLERS 406 

158.  CONTACT  or  CONCENTRATED  LOADS 407 

159.  CIRCULAR  PLATES  WITH  UNIFORM  LOAD 409 

160.  CIRCULAR  PLATES  WITH  CONCENTRATED  LOAD 411 

161.  ELLIPTICAL  PLATES 414 

162.  RECTANGULAR  PLATES 415 

163.  HOLLOW  SPHERES 417 

CHAPTER  XVIII 
MISCELLANEOUS  DISCUSSIONS 

ART.  164.  CENTRIFUGAL  TENSION 421 

165.  CENTRIFUGAL  FLEXURE 425 

166.  UNSYMMETRIC  LOADS  ON  BEAMS 427 

167.  INFLUENCE  LINES 431 

168.  CURVED  BEAMS 433 

169.  PRODUCT  OF  INERTIA 437 

170.  MOMENT  OF  INERTIA 438 

171.  SPRINGS 442 

CHAPTER  XIX 
MATHEMATICAL  THEORY  OF  ELASTICITY 

ART.  172.    INTRODUCTION 447 

173.  ELASTIC  CHANGES  IN  VOLUME     .     .      , 448 

174.  NORMAL  AND  TANGENTIAL  STRESSES      ........  450 

175.  RESULTANT  STRESSES ......  452 

176.  THE  ELLIPSOID  OF  STRESS      ...........  454 

177.  THE  THREE  PRINCIPAL  STRESSES      .  455 

178.  MAXIMUM  SHEARING  STRESSES     ..••••••..  457 

179.  DISCUSSION  OF  A  CRANK  PIN      .     .     ,     „     .     ,     .     .     „     .  459 


CONTENTS  xi 

PAGE 

180.  THE  ELLIPSE  OF  STRESS 461 

181.  SHEARING  MODULUS  OF  ELASTICITY. 463 

182.  THE  VOLUMETRIC  MODULUS   ...... 465 

183.  STORED  INTERNAL  ENERGY 467 

CHAPTER  XX 
TESTING  OF  MATERIALS 

ART.  184.  TESTING  MACHINES 470 

185.  TEST  SPECIMENS 474 

186.  TENSILE  TESTS 476 

187.  COMPRESSIVE  TESTS 478 

188.  MISCELLANEOUS  TESTS 479 

189.  SPECIFICATIONS  FOR  STRUCTURAL  STEEL 482 

190.  UNIFORMITY  IN  TESTING 486 

APPENDIX 

ART.  191.  VELOCITY  OF  STRESS        488 

192.  ELASTIC-ELECTRIC  ANALOGIES 490 

193.  MISCELLANEOUS  PROBLEMS 491 

194.  ANSWERS  TO  PROBLEMS         ......  493 

196.  EXPLANATION  OF  TABLES 494 

TABLES 

TABLE    1.  AVERAGE  WEIGHT  AND  EXPANSIBILITY 496 

2.  AVERAGE  ELASTIC  PROPERTIES 496 

3;  AVERAGE  TENSILE  AND  COMPRESSIVE  STRENGTH       .     .     .     .497 

4.  AVERAGE  SHEARING  AND  FLEXURAL  STRENGTH 497 

5.  WORKING  UNIT-STRESSES  FOR  BUILDINGS 498 

6.  STEEL  I-BEAM  SECTIONS 499 

7.  STEEL  BULB-BEAM  SECTIONS 500 

8.  STEEL  T  SECTIONS 500 

9.  STEEL  CHANNEL  SECTIONS     . 501 

10.  STEEL  ANGLE  SECTIONS 502 

11.  STEEL  Z  SECTIONS 503 

12.  COMPARISON  OF  BEAMS „ 503 

13.  GERMAN  I  BEAMS 504 

14.  WEIGHT  OF  WROUGHT-!RON  BARS 505 

15.  SQUARES  OF  NUMBERS 506 

16.  AREAS  OF  CIRCLES 508 

17.  TRIGONOMETRIC  FUNCTIONS 510 

18.  LOGARITHMS  OF  TRIGONOMETRIC  FUNCTIONS 511 

19.  LOGARITHMS  OF  NUMBERS      .     .     .     .     .     .     .     .     .     .     .512 

20.  CONSTANTS  AND  THEIR  LOGARITHMS     ....     0     ...  514 

INDEX    .     .     o 515 


MECHANICS  OF  MATERIALS 

CHAPTER  I 
ELASTIC  AND  ULTIMATE  STRENGTH 

ARTICLE  1.    SIMPLE  AXIAL  STRESSES 

Mechanics  of  Materials  is  the  science  that  treats  of  the  effects 
of  forces  in  causing  changes  in  the  size  and  shape  of  bodies.  Such 
forces  are  generally  applied  to  bodies  slowly,  and  the  changes  in 
size  and  shape  occur  while  the  forces  are  increasing  up  to  their 
final  values.  A  '  Stress '  is  an  internal  force  that  resists  the  change 
in  shape  or  size,  and  when  the  applied  forces  have  reached  their 
final  values  the  internal  stresses  hold  them  in  equilibrium.  The 
simplest  case  is  that  of  a  rope,  at  each  end  of  which  a  man  pulls 
with  a  force,  say  25  pounds,  then  in  every  section  of  the  rope 
there  exists  a  stress  of  25  pounds.  Stresses  are  measured  in  the 
same  unit  as  that  used  for  the  applied  forces,  and  generally  in 
pounds  or  kilograms. 

A  'Bar'  is  a  prismatic  body  having  the  same  size  throughout 
its  length.  If  a  plane  is  passed  normal  to  the  bar,  its  intersection 
with  the  prism  is  called  the  'cross-section'  or  the  'section'  of  the 
bar,  and  the  area  of  this  cross-section  is  called  the  '  section  area. ' 
In  any  section  imagined  to  be  cut  out,  there  exists  a  stress  equal 
to  the  longitudinal  force  acting  on  the  end  of  the  bar.  A  'Unit- 
Stress'  is  the  stress  on  a  unit  of  the  section  area,  and  this  is  usually 
expressed  in  pounds  per  square  inch  or  in  kilograms  per  square 
centimeter.  For  example,  let  a  bar,  3  inches  wide  and  ij  inches 
thick,  be  subjected  to  a  pull  of  14  400  pounds;  the  resisting  stress 
is  14  400  pounds,  and  the  unit-stress  is  14  400  pounds  divided  by 
4j  square  inches,  or  3  200  pounds  per  square  inch. 


ELASTIC  AND  ULTIMATE  STRENGTH 


CHAP.  I 


When  external  forces  act  upon  the  ends  of  a  bar  in  a  direction 
away  from  its  ends  they  are  called  'Tensile  Forces';  when  they 
act  towards  the  ends,  they  are  called  * Compressive  Forces.'  A 
pull  is  a  tensile  force  and  a  push  is  a  compressive  force,  and 
these  two  cases  are  frequently  called  'Tension'  and  'Compres- 
sion'. The  resisting  stresses  receive  similar  designations;  a  ten- 
sile stress  is  that  which  resists  tensile  forces;  a  compressive  stress 
is  that  which  resists  compressive  forces. 


im 


Fig.  la  Fig.  Ib 

The  case  of  tension  is  shown  in  Fig.  la,  where  two  tensile 
forces,  each  equal  to  P,  act  upon  the  ends  of  a  bar  having  the 
section  area  a.  Let  mn  be  any  imaginary  plane  normal  to  the 
bar,  and  let  the  two  parts  of  the  bar  be  imagined  to  be  separated 
as  in  Fig.  Ib.  Then  the  equilibrium  of  each  part  will  be  main- 
tained if  tensile  forces  equivalent  to  the  resisting  stresses  are 
applied  as  shown.  These  resisting  stresses  act  normally  to  the 
section  area  a,  and  they  are  in  each  case  opposite  in  direction  to 
the  force  P.  Each  part  of  the  bar  is  held  in  equilibrium  by  the 
applied  force  P  and  the  resisting  tensile  stress;  accordingly  the 
resisting  tensile  stress  must  equal  the  tensile  force  P. 


im 


Fig.  Ic  Fig.  Id 

The  case  of  compression  is  shown  in  Fig.  Ic,  where  the  forces 
P  act  toward  the  ends  of  the  bar.  For  any  imaginary  plane  mn, 
the  bar  may  be  regarded  as  separated  into  two  bars  as  in  Fig. 
Id,  each  of  which  is  held  in  equilibrium  by  compressive  stresses 
acting  normally  to  the  section  area  and  in  directions  opposite  to 
P.  The  total  resisting  compressive  stress  must  be  equal  to  P 
in  order  that  equilibrium  may  prevail. 


.  1  SIMPLE  AXIAL  STRESSES  3 

Let  S  be  the  unit-stress  of  tension  or  compression,  as  the  case 
may  be,  which  acts  in  any  normal  section  of  a  bar  having  the 
area  a.  The  total  stress  on  the  section  is  then  Sa,  and  this  is 
uniformly  distributed  over  the  area  a  when  the  force  P  acts 
along  the  axis  of  the  bar;  then, 

Sa=P  S=P/a  a=P/S  (1) 

from  which  one  of  the  quantities  may  be  computed  when  the 
other  is  given.  For  example,  let  it  be  required  to  find  what  the 
section  area  of  a  stick  of  timber  should  be  when  it  is  subject  to 
a  pull  of  1 6  500  pounds,  it  being  required  that  the  tensile  unit- 
stress  shall  be  900  pounds  per  square  inch;  here  a=  16  500/900  = 
18.3  square  inches. 

The  terms  'Axial  Forces'  and  'Axial  Stresses'  are  used  to 
include  both  tension  and  compression  acting  upon  a  bar,  it  being 
understood  that  the  resultant  of  the  applied  forces  acts  along  the 
axis  of  the  bar.  The  axial  force  P  is  often  called  a  'Load'.  It 
is  always  understood,  unless  otherwise  stated,  that  the  stresses 
due  to  an  axial  load  are  uniformly  distributed  over  the  section 
area,  and  this  is  called  the  case  of  '  Simple  Axial  Stress ',  it  being 
one  of  the  most  common  cases  in  engineering.  Cases  where  the 
stress  is  not  uniformly  distributed  over  the  section  area  occur 
when  the  resultant  of  the  applied  forces  does  not  act  along  the 
axis  of  the  bar,  and  also  in  beams  and  long  columns. 

The  first  effect  of  an  axial  load  is  to  change  the  length  of  the 
bar  upon  which  it  acts.  This  'Deformation'  continues  until  the 
resisting  stresses  have  attained  such  magnitudes  that  they  equili- 
brate the  applied  forces.  The  deformation  of  a  bar  which  occurs 
in  tension  is  called  'Elongation',  and  that  which  occurs  in  com- 
pression is  called  'Shortening'.  As  the  applied  forces  increase, 
the  resisting  stresses  also  increase,  until  finally  the  resistance  is 
unable  to  balance  the  force,  the  deformation  rapidly  increases, 
and  the  bar  breaks  or  ruptures.  The  above  equations  apply 
also  to  the  case  of  rupture.  For  example,  it  is  known  that  a 
cast-iron  bar  will  rupture  under  tension  when  the  unit-stress  S 
becomes  about  20  ooo  pounds  per  square  inch ;  if  the  bar  is 
inches  in  cross-section,  its  section  area  is  if  square 


4  ELASTIC  AND  ULTIMATE  STRENGTH  CHAP.  I 

inches,    and    the    tensile    force    required    to    cause    rupture    is 
P  =  if  X  20  ooo  =  37  500  pounds. 

Problem  la.  If  a  cast-iron  bar,  1^X2  inches  in  section,  breaks 
under  a  tensile  load  of  60  ooo  pounds,  what  load  will  break  a  cast- 
iron  rod  of  if  inches  diameter? 

Prob.  Ib.  A  cast-iron  bar  which  is  to  be  subjected  to  a  tension  of 
34  ooo  pounds  is  to  be  designed  so  that  the  unit-stress  shall  be  2  500 
pounds  per  square  inch.  If  the  bar  is  round  what  should  be  its 
diameter  ? 

ART.  2.     THE  ELASTIC  LIMIT 

When  a  bar  is  subjected  to  a  gradually  increasing  tension,  the 
bar  elongates,  and  up  to  a  certain  limit  it  is  found  that  the  elon- 
gation is  proportional  to  the  load.  Thus,  when  a  bar  of  wrought 
iron  one  square  inch  in  section  area  and  100  inches  long  is  sub- 
jected to  a  load  of  5  ooo  pounds,  it  is  found  to  elongate  closely 
0.02  inches;  when  10  ooo  pounds  is  applied,  the  total  elongation 
is  0.04  inches;  when  15  ooo  pounds  is  applied,  the  elongation  is 
0.06  inches;  when  20000  pounds'  is  applied,  the  elongation  is 
0.08  inches;  when  25  ooo  pounds  is  applied,  the  elongation  is 
o.io  inches.  Thus  far,  each  addition  of  5  ooo  pounds  has  pro- 
duced an  additional  elongation  of  0.02  inches.  But  when  the 
next  5  ooo  pounds  is  added,  making  a  total  load  of  30  ooo  pounds, 
it  is  found  that  the  total  elongation  is  about  half  an  inch,  and 
hence  the  elongations  are  increasing  in  a  faster  ratio  than  the 
applied  loads  and  the  resisting  stresses. 

The  'Elastic  Limit'  is  denned  to  be  that  unit-stress  at  which 
the  deformation  begins  to  increase  in  a  faster  ratio  than  the 
applied  loads.  In  the  above  example  this  limit  is  about  25  ooo 
pounds  per  square  inch,  and  this  is  the  average  value  of  the  elastic 
limit  for  wrought  iron.  Instead  of  Elastic  Limit  the  terms  '  Pro- 
portional Elastic  Limit '  and  '  Proportional  Limit '  are  often  used. 

When  the  unit-stress  in  a  bar  is  not  greater  than  the  elastic 
limit,  the  bar  returns,  on  the  removal  of  the  load,  to  its  original 
length.  Thus,  the  above  wrought-iron  bar  was  100.10  inches  long 
under  the  load  of  25  ooo  pounds,  and  on  the  removal  of  that  load 


ART.  2  THE  ELASTIC  LIMIT  5 

it  returns  to  its  original  length  of  100.00  inches.  When  the  unit- 
stress  is  greater  than  the  elastic  limit,  the  bar  does  not  fully  re- 
turn to  its  original  length,  but  there  remains  a  so-called  *  Perma- 
nent Set'.  For  instance  let  the  length  of  the  above  bar  under  a 
stress  of  34  ooo  pounds  be  102  inches,  and  on  the  removal  of  the 
tension  let  its  length  be  loif  inches;  then  the  permanent  set  of 
the  bar  is  ij  inches. 

In  all  cases  of  simple  axial  tension  the  resisting  stress  is  equal 
to  the  load,  and  the  stresses  hence  increase  proportionately  to  the 
loads.  When  the  elastic  limit  is  not  exceeded,  the  elongations 
are  found  to  be  proportional  to  the  loads,  but  when  this  limit  is 
exceeded  they  increase  faster  than  the  loads,  and  a  permanent  set 
remains.  Therefore  the  elastic  properties  of  a  bar  are  injured 
when  it  is  stressed  beyond  the  elastic  limit.  Accordingly  it  is  a 
fundamental  rule  in  designing  engineering  constructions  that  the 
unit-stresses  should  not  exceed  the  elastic  limit  of  the  material. 

The  above  facts  regarding  the  behavior  of  materials  in  ten- 
sion have  been  ascertained  by  many  tests  of  bars  and  are  to  be 
regarded  as  fundamental  laws;  all  experience  and  all  experi- 
ments have  verified  these  laws  as  being  approximately  true  for 
the  common  materials  used  in  engineering.  By  such  tests  also  it 
has  been  shown  that  such  laws  apply  to  compression  as  well  as 
to  tension.  The  following  are  approximate  average  values  of  the 
elastic  limits  in  tension  for  five  materials  extensively  used  in  engi- 
neering construction: 

Material  Elastic  Limit 

Timber  3  ooo  pounds  per  square  inch 

Cast  Iron  6  ooo  pounds  per  square  inch 

Wrought  Iron  25  ooo  pounds  per  square  inch 

Structural  Steel  35  ooo  pounds  per  square  inch 

Strong  Steel  50  ooo  pounds  per  square  inch 

These  values  should  be  carefully  memorized  by  the  student,  and 
be  used  in  the  solution  of  the  problems  in  the  following  pages. 
Table  2,  at  the  end  of  this  volume,  gives  these  constants  in  the 
metric  system  of  measures. 

The  above  average  values  are  subject  to  considerable  varia- 


6  ELASTIC  AND  ULTIMATE  STRENGTH  CHAP.  I 

tion  for  different  qualities  of  the  same  material;  for  example, 
some  grades  of  structural  steel  may  have  an  elastic  limit  ten  per- 
cent lower  than  35  ooo  pounds  per  square  inch,  while  others  may 
run  ten  percent  higher.  In  testing  a  number  of  bars  of  the  same 
kind,  indeed,  it  is  not  uncommon  to  find  a  variation  of  five  or 
ten  percent  in  the  different  results. 

The  elastic  limit  in  compression  is  the  same  as  that  in  tension 
for  all  the  above  materials  except  cast  iron,  which  has  about  three 
times  the  value  above  given.  Brittle  materials,  like  brick,  stone, 
and  cast  iron,  usually  have  higher  elastic  limits  in  compression 
than  in  tension,  but  it  will  be  seen  later  that  the  elastic  limits  for 
such  materials  are  poorly  defined,  that  is  it  is  difficult  to  determine 
them  with  exactness. 

By  the  help  of  the  above  experimental  values  and  the  princi- 
ples of  Art.  1,  many  simple  problems  in  investigation  and  design 
may  be  solved.  For  example,  let  it  be  required  to  find  the  size 
of  a  square  stick  of  timber  to  carry  a  compressive  load  of  64  ooo 
pounds,  so  that  the  unit-stress  may  be  one-third  of  the  elastic 
limit;  here  the  elastic  limit  is  3  ooo  pounds  per  square  inch,  and 
the  section  area  required  is  64  ooo/i  ooo,  or  64  square  inches,  so 
that  a  stick  8x8  inches  in  size  is  needed. 

Prob.  20.  Find  the  diameter  of  a  round  rod  of  wrought  iron,  which 
is  to  be  under  a  tension  of  64  ooo  pounds,  so  that  the  unit-stress  may 
be  one-third  of  the  elastic  limit. 

Prob.  2b.  A  stick  of  timber  3  inches  thick  is  under  a  tension  of 
1 2  ooo  pounds.  Compute  its  width,  so  that  the  unit-stress  may  be  40 
percent  of  the  elastic  limit. 

ART.  3.    ULTIMATE  STRENGTH 

When  the  section  area  of  a  bar  is  under  an  axial  unit-stress 
exceeding  the  elastic  limit  of  its  material,  the  bar  is  usually  in 
an  unsafe  condition.  As  the  external  forces  increase,  the  defor- 
mation increases  in  a  more  rapid  ratio,  until  finally  the  rupture 
of  the  bar  occurs.  The  term  'ultimate  strength'  is  used  to  desig- 
nate the  highest  unit-stress  that  the  bar  can  sustain,  this  occur- 
ring at  or  just  before  rupture. 


ART.  3  ULTIMATE  STRENGTH  7 

The  ultimate  strengths  of  materials  are  usually  from  two 
to  four  times  their  elastic  limits,  and  for  some  materials  they 
are  much  higher  in  compression  than  in  tension.  Thus,  the 
ultimate  tensile  strength  of  cast  iron  is  about  20  ooo  pounds  per 
square  inch,  while  its  ultimate  compressive  strength  is  about 
90  ooo  pounds  per  square  inch. 

Average  values  of  the  ultimate  strengths  of  materials  are 
given  in  the  following  articles,  but  these  are  subject  to  much 
variation  for  different  qualities  of  materials.  Thus,  inferior 
grades  of  cast  iron  may  have  a  tensile  strength  as  low  as  15  ooo 
pounds  per  square  inch,  while  the  best  grades  are  often  higher 
than  30  ooo  pounds  per  square  inch.  In  general  a  variation  of 
ten  percent  from  these  average  values  is  to  be  regarded  as  liable 
to  occur. 

The  'Factor  of  Safety '  of  a  bar  under  stress  is  the  number 
which  results  by  dividing  the  ultimate  strength  of  the  material 
by  the  actual  unit-stress  on  the  section  area.  For  example,  let 
a  stick  of  timber,  6X6  inches  in  section  area,  be  under  a  ten- 
sion of  32  400  pounds.  The  actual  unit-stress  is  then  32  400/36 
=  900  pounds  per  square  inch.  Since  the  average  tensile  strength 
of  timber  is  about  10  ooo  pounds  per  square  inch,  the  factor  of 
safety  of  the  bar  is  10  000/900=1  n. 

The  factor  of  safety  was  formerly  much  used  in  designing, 
and  for  timber  under  steady  tension  was  taken  as  about  10;  that 
is,  one- tenth  of  the  ultimate  strength  was  regarded  as  the  highest 
allowable  unit-stress.  By  this  method,  timber  having  an  ulti- 
mate tensile  strength  of  12  ooo  pounds  per  square  inch  should 
be  subjected  to  a  unit-stress  of  only  12000/10=1  200  pounds 
per  square  inch,  so  that  the  section  area  of  a  stick  under  a  ten- 
sion of  19  200  pounds  should  be  19  200/1  200  =  16  square  inches. 

It  is  now  considered  a  better  plan,  in  judging  of  the  degree 
of  security  of  a  body  under  stress,  to  consider  the  elastic  limit 
of  the  material.  Thus,  for  a  stick  of  timber,  the  elastic  limit 
is  about  3  ooo  pounds  per  square  inch,  and  the  actual  unit-stress 
in  tension  should  be  less  than  this,  say  one-half  or  one-third, 
according  as  the  applied  forces  are  steady  or  variable.  The 


8  ELASTIC  AND  ULTIMATE  STRENGTH  CHA>.  j 

method  of  the  factor  of  safety  is,  however,  of  much  value  to  a 
student  and  it  will  be  often  used  in  this  volume.  In  practice 
both  ultimate  strengths  and  elastic  limits  must  be  considered 
in  deciding  upon  the  allowable  unit-stresses  to  which  the  parts 
of  engineering  constructions  are  to  be  subjected.  The  average 
values  of  the  ultimate  strengths  of  the  materials  of  engineering 
are  tabulated  at  the  end  of  this  volume  in  both  English  anc? 
metric  measures. 

The  theoretical  and  experimental  conclusions  thus  far  estab- 
lished, regarding  the  behavior  of  a  bar  under  tension  or  com- 
pression, may  now  be  formulated  in  the  following  laws: 

1.  The  resisting  axial  stress  in  every  section  of  a  bar  is 
equal  to  the  applied  tensile  or  compressive  load. 

2.  Under  small  stresses,  the  deformations  of  the  bar  are 
proportional  to  the  loads,  and  hence  also  to  the  unit-stresses. 
When  the  unit-stress  is  less  than  the  value  called  the  elastic 
limit,  the   bar   springs   back   to    its    original  length   on  the 
removal  of  the  load. 

3.  When  the  unit-stress  is  greater  than  the  elastic  limit, 
the  deformations  increase  in  a  faster  ratio  than  the  loads  and 
stresses,  and  the  bar  does  not  spring  back  to  its  original 
length  on  the  removal  of  the  load. 

4.  When  the  load  becomes  sufficiently  great,  the  resisting 
stress  fails  to  balance  it,  so  that  the  deformation  rapidly  in- 
creases and  the  bar  ruptures. 

5.  The  allowable  unit-stresses  used  in  engineering  prac- 
tice are  less  than  the  elastic  limit  of  the  material. 

The  first  of  these  laws  is  a  theoretical  one  and  rigidly  correct 
for  all  cases,  it  being  in  fact  a  particular  case  of  Newton's  law 
that  action  and  reaction  are  equal  and  opposite.  The  second 
law  applies  strictly  only  to  elastic  materials  like  wrought  iron 
and  steel,  and  is  only  roughly  applicable  to  brittle  materials 
like  stone  and  cast  iron.  The  third  law  applies  to  all  kinds 
of  materials,  but  for  brittle  ones  the  elastic  limit  is  difficult 
to  determine. 

Prob.  30.  A  bar  of  structural  steel,  2j  inches  in  diameter,  ruptures 
under  a  tension  of  271  ooo  pounds.  What  is  the  ultimate  tensile 
strength  ? 


ART.  4 


TENSION 


9 


Prob.  36.  Using  the  value  given  above  for  the  ultimate  tensile 
strength  of  cast  iron,  compute  the  tensile  force  which  is  required  to 
rupture  a  cast-iron  bar  which  is  2jX  if  inches  in  section  area. 


ART.  4.    TENSION 

A  tensile  test  of  a  vertical  bar  may  be  made  by  fastening 
its  upper  end  firmly  with  clamps  and  then  applying  loads  to 
its  lower  end.  The  elongations  of  the  bar  are  found  to  increase 
proportionately  to  the  loads,  and  hence  also  to  the  internal  ten- 
sile stresses,  until  the  elastic  limit  of  the  material  is  reached 
(Art.  2).  After  the  unit-stress  has  exceeded  the  elastic  limit, 
the  elongations  increase  more  rapidly  than  the  loads,  and  this 
is  often  accompanied  by  a  reduction  in  area  of  the  cross-section 
of  the  bar.  Finally,  the  ultimate  tensile  strength  of  the  mate- 
rial is  reached  and  the  bar  breaks. 


80000 


0.02 


0.10 


0.18 


0.04  0.06  0.08 

Elongations  per  unit  of  length 

Fig.  4 

A  graphical  illustration  of  these  phenomena  may  be  made 
by  laying  off  the  unit-stresses  as  ordinates  and  the  elongations 
per  unit  of  length  as  the  abscissas.  At"  various  intervals,  as  the 
test  progresses,  the  applied  loads  are  measured  and  also  the 
corresponding  elongations.  The  load  divided  by  the  section 
area  of  the  bar  gives  the  unit-stress,  while  the  total  elongation 
divided  by  the  length  of  the  bar  gives  the  unit-elongation.  .On 


10  ELASTIC  AND  ULTIMATE  STRENGTH  CHAP.  I 

the  diagram  a  point  is  put  at  the  intersection  of  each  unit-stress 
with  its  corresponding  unit-elongation,  and  a  curve  is  drawn 
connecting  the  plotted  points.  Fig.  4  represents  mean  curves 
obtained  in  this  manner  for  timber,  cast  iron,  wrought  iron,  and 
steel.  It  is  seen  that  each  curve  is  a  straight  line  from  the  origin 
until  the  elastic  limit  is  reached,  showing  that  the  unit-elonga- 
tions increase  proportionally  to  the  unit-stresses.  At  the  elastic 
limit  a  sudden  change  in  the  curve  is  seen,  and  afterwards  the 
elongation  increases  more  rapidly  than  the  stress.  This  dia- 
gram gives  mean  comparative  curves  only,  and  the  curve  for 
any  individual  test  might  deviate  considerably  from  that  shown. 

The  end  of  the  curve  marks  the  rupture  of  the  bar.  For 
example,  it  is  seen  that  timber  ruptures  in  tension  under  a  stress 
of  about  10  ooo  pounds  per  square  inch,  and  that  the  elongation 
of  the  bar  at  that  time  is  about  i-J$-0-  of  the  length  of  the  bar, 
that  is  the  total  elongation  is  about  1.5  percent  of  the  length.  The 
rupture  of  wrought  iron  and  structural  steel  is  not  shown  upon 
the  diagram,  since  the  ultimate  elongation  of  these  materials  is 
about  30  percent,  while  the  diagram  extends  only  up  to  12  per- 
cent. The  following  are  approximate  average  values  of  the 
ultimate  strengths  and  ultimate  elongations  for  five  materials 
widely  used  in  engineering  work: 

Material  Ultimate  Tensile  Strength  Ultimate  Elongation 

Timber  10  ooo  pounds  per  square  inch        1.5  percent 

Cast  Iron  20  ooo  pounds  per  square  inch       0.3  percent 

Wrought  Iron  50  ooo  pounds  per  square  inch  30  percent 
Structural  Steel  60  ooo  pounds  per  square  inch  30  percent 
Strong  Steel  100  ooo  pounds  per  square  inch  15  percent 

These  ultimate  tensile  strengths  should  be  carefully  kept  in 
mind  by  the  student  as  a  basis  for  future  knowledge,  and  they 
will  be  used  in  the  solution  of  the  examples  and  problems  in 
this  book.  Table  3,  at  the  end  of  this  volume,  gives  values 
for  other  materials. 

These  average  values  of  ultimate  strengths  and  elongations 
are  those  derived  from  tests  on  small  specimens,  say  one  inch 
in  diameter  and  8  inches  in  gaged  length.  Large  bars  such  as 


A.RT.  5  COMPRESSION  11 

are  used  in  engineering  constructions  have  ultimate  strengths 
slightly  less  and  ultimate  elongations  considerably  less  than  these 
values.  One  of  the  reasons  for  using  factors  of  safety  (Art.  3) 
is  to  provide  security  against  the  smaller  ultimate  strength  of 
the  large  pieces  which  must  necessarily  be  used.  The  ultimate 
elongation  is  an  index  of  the  toughness  and  ductility  of  the  mate- 
rial, but  it  is  not  used  in  the  computations  of  designing.  The 
unit-elongation  which  occurs  when  a  bar  is  stressed  up  to  the 
elastic  limit  is  very  small,  compared  with  that  which  obtains  at 
rupture,  as  the  curves  in  Fig.  4  plainly  show.  Fig.  185&  shows 
two  steel  specimens  that  were  ruptured  by  tension. 

Steel  is  a  material  which  has  variable  physical  properties 
depending  upon  its  chemical  composition  and  the  method  of 
its  manufacture.  'Structural  steel'  is  that  which  is  used  in 
buildings,  bridges,  and  ships,  and  it  resembles  wrought  iron  in 
general  behavior,  but  its  tensile  strength  is  about  twenty  percent 
higher.  '  Strong  steel '  is  not  a  trade  term  and  is  here  introduced 
for  instruction  purposes  only.  The  grades  of  steel  are  very 
numerous,  and  they  range  in  tensile  strength  from  50  ooo  to 
250000  pounds  per  square  inch.  The  terms  'soft'  and  'hard' 
are  often  used  to  designate  steel  with  low  and  high  tensile  strengths. 

Prob.  4a.  A  steel  specimen,  0.505  inches  in  diameter,  reached  the 
elastic  limit  under  a  tensile  load  of  7800  pounds  and  ruptured  under 
a  load  of  14  800  pounds.  Compute  the  elastic  limit  and  the  ultimate 
strengths  of  the  steel. 

Prob.  46.  This  specimen  had  a  length  of  2.00  inches  between  two 
marks  made  on  it  before  the  test.  At  the  elastic  limit  the  distance  be- 
tween these  two  marks  was  2.003  inches  and  after  rupture  it  was  2.45 
inches.  Compute  the  unit-elongation  for  the  elastic  limit  and  for 
rupture. 

ART.  5.     COMPRESSION 

The  phenomena  of  compression  are  the  reverse  of  those  of 
tension  in  regard  to  the  direction  of  the  applied  forces  and  resist- 
ing stresses.  When  loads  are  applied  to  compress  a  bar,  the 
amount  of  shortening  is  proportional  to  the  load,  provided  the 
unit-stress  on  the  material  does  not  exceed  the  elastic  limit.  After 


12  ELASTIC  AND  ULTIMATE  STRENGTH  CHAP.  I 

the  elastic  limit  is  passed,  the  shortening  increases  more  rapidly 
than  the  load  and  hence  more  rapidly  than  the  unit-stress,  and 
finally  the  rupture  of  the  bar  takes  place.  The  area  of  the  cross- 
section  decreases  under  tension  and  increases  under  compression. 

The  simplest  way  of  testing  a  bar  by  compression  is  to  place 
it  on  a  firm  foundation  and  put  the  load  upon  its  upper  face 
as  shown  in  Fig.  5a.  The  bar  is  held  in  equilibrium  by  the 
load  P  and  the  equal  upward  reaction  of  the  support  and  there 
may  be  represented  by  arrows  as  in  Fig.  5b.  On  any  horizontal 
section  of  the  bar,  there  are  acting  compressive  stresses  the  sum 
of  which  equals  P  (Art.  1).  If  the  section  area  is  a  the  average 
compressive  unit-stress  is  P/a,  and  this  will  be  uniformly  dis- 
tributed over  the  section  when  the  forces  P  coincide  in  direc- 
tion with  the  axis  of  the  prism. 


Fig.  5a  Fig.  5b  Fig.  5$  Fig.  5d 

When  the  length  of  the  bar  does  not  exceed  about  ten  times 
its  least  lateral  dimension,  rupture  sometimes  occurs  by  an  obliqu.3 
splitting  or  shearing  as  shown  in  Fig.  5c.  When  the  length  is 
large  compared  with  the  thickness,  failure  occurs  under  a  side- 
wise  bending,  as  seen  in  Fig.  5d.  The  short  specimens  are  cases 
of  simple  compression,  and  the  values  given  in  the  following 
table  refer  only  to  these;  the  long  specimens  are  called  'columns' 
and  in  them  other  stresses  are  developed  than  that  of  simple 
compression  (Art.  77).  In  general,  the  term  'compressive 
strength'  refers  to  that  obtained  from  a  bar  the  length  of  which 
is  considerably  less  than  ten  times  its  thickness.  Fig.  185c 
shows  the  rupture  of  a  cement  cube  and  a  timber  block. 

Graphical  illustrations  of  the  behavior  of  materials  under 
compressive  stress  may  be  made  in  the  same  manner  as  for  ten- 
sion, the  unit-shortenings  being  laid  off  as  abscissas.  In  most 
cases  the  ultimate  shortening  is  much  less  than  the  ultimate 


ART.  5  COMPKESSION  13 

elongation.  For  steel  the  elastic  limit  is  about  the  same  in  com- 
pression as  in  tension,  but  the  compressive  strength  of  hard 
steel  is  much  higher  than  the  tensile  strength.  The  following 
are  average  values  of  the  ultimate  compressive  strengths  of 
the  principal  materials  used  in  engineering  construction: 

Material  Ultimate  Compressive  Strength 

Brick  3  ooo  pounds  per  square  inch 

Stone  6  ooo  pounds  per  square  inch 

Timber  8  ooo  pounds  per  square  inch 

Cast  Iron  90  ooo  pounds  per  square  inch 

Wrought  Iron  50  ooo  pounds  per  square  inch 

Structural  Steel  60  ooo  pounds  per  square  inch 

Strong  Steel  120  ooo  pounds  per  square  inch 

These  ultimate  strengths  are  subject  to  much  variation  for 
different  qualities  of  materials,  but  it  is  necessary  for  the  student 
to  fix  them  in  his  mind  as  a  basis  for  future  knowledge.  It  is 
seen  that  timber  is  not  quite  as  strong  in  ultimate  compression 
as  in  tension,  that  cast  iron  is  4^  times  as  strong,  that  wrought 
iron  and  structural  steel  have  the  same  strength  in  the  two  cases, 
and  that  the  typical  material  called  strong  steel  has  a  higher 
strength  in  compression  than  in  tension.  There  are  several 
varieties  of  hard  steel  which  have  ultimate  compressive  strengths 
much  greater  than  that  above  given  (Art.  25). 

The  investigation  of  a  body  under  compression  is  made  in 
the  same  manner  as  for  one  under  tension.  For  example,  if 
a  stone  block,  8X12  inches  in  cross-section,  is  subject  to  a  com- 
pression of  230000  pounds  the  unit-stress  is  230000/96  =  2400 
pounds  per  square  inch,  and  the  factor  of  safety  is  6  000/2  400  = 
2j;  this  is  not  sufficiently  high  for  stone,  as  will  be  seen  in  Art.  7. 

Prob.  5#.  A  solid  cast-iron  cylinder  of  4  inches  diameter  is  under 
a  compression  of  600  ooo  pounds.  Compute  its  factor  of  safety. 

Prob.  5b.  A  brick  2X4X8  inches  weighs  about  4^  pounds.  What 
must  be  the  height  of  a  pile  of  bricks  so  that  the  compressive  unit- 
stress  on  the  lowest  brick  shall  be  one-half  of  its  ultimate  strength  ? 

Prob.  5c.  A  bar  of  wrought  iron  one  square  inch  in  section  area 
and  one  yard  long  weighs  10  pounds.  Find  the  length  of  a  vertical 
bar  so  that  the  stress  at  the  upper  end  shall  equal  the  elastic  limit. 


14  ELASTIC  AND  ULTIMATE  STRENGTH  CHAP,  i 

ART.  6.    SHEAR 

When  two  equal  and  opposite  forces  act  at  right  angles  to 
a  bar  and  are  very  near  together,  they  are  called  l  Shearing  Forces  ', 
and  they  tend  to  cut  or  shear  the  bar.  The  action  of  the  forces 
is  similar  to  those  in  a  pair  of  shears,  from  which  analogy  the 
name  is  derived,  and  the  resisting  stresses  are  called  'shearing 
stresses '.  Tension  and  compression  cause  stresses  which  act 
normally  to  a  section  area,  but  shear  causes  stresses  which  act 
parallel  with  and  along  the  section  area.  Unless  otherwise  stated 
it  is  to  be  considered  that  the  shearing  stresses  generated  by  shear- 
ing forces  are  uniformly  distributed  over  the  section  area. 


of 

Fig.  6a  Fig.  66 

Fig.  6a  shows  the  case  of  two  plates  fastened  together  with 
a  rivet  and  subject  to  a  tension  P.  Let  the  section  area  of  the 
rivet  be  a;  then  the  shearing  unit-stress  on  that  cross-section 
lying  in  the  plane  where  the  plates  overlap  is  P/a,  and  if  this 
equals  the  ultimate  shearing  strength  of  the  material,  the  rivet 
will  rupture  by  shearing.  Fig.  Qb  shows  the  case  of  a  beam 
resting  upon  two  supports  and  carrying  two  equal  loads  P  near 
the  supports,  the  reaction  of  each  support  being  P;  here  a 
resisting  shearing  stress  equal  to  P  acts  on  each  side  of  the 
section  mn,  the  stress  on  the  left  of  mn  acting  downward  and 
that  on  the  right  acting  upward;  in  this  case  also  the  shear- 
ing unit-stress  is  P/a,  and  the  bar  will  shear  off  when  this  is 
equal  to  the  ultimate  shearing  strength  of  the  material. 

The  following  are  average  values  of  the  ultimate  shearing 
strengths  of  materials  as  determined  by  experiment: 

Material  Ultimate  Shearing  Strength 

Brick  i  ooo  pounds  per  square  inch 

Stone  i  500  pounds  per  square  inch 

Timber,  along  grain  500  pounds  per  square  inch 

Timber,  across  grain  3  ooo  pounds  per  square  inch 

Cast  Iron  18  ooo  pounds  per  square  inch 


ART.  6  SHEAR  15 

Material  Ultimate  Shearing  Strength 

Wrought  Iron  40  ooo  pounds  per  square  inch 

Structural  Steel  50  ooo  pounds  per  square  inch 

Strong  Steel  80  ooo  pounds  per  square  inch 

By  comparing  these  with  the  values  for  tension  in  Art.  4, 
it  is  seen  that  the  shearing  strength  of  timber  is  about  one-third 
of  the  tensile  strength  when  the  shearing  occurs  across  the  grain, 
and  very  much  smaller  when  it  occurs  parallel  with  the  grain. 
For  cast  iron  the  shearing  strength  is  about  90  percent,  and  for 
wrought  iron  and  steel  it  is  about  80  percent  of  the  tensile  strength. 

There  is  an  elastic  limit  in  shear  as  well  as  in  tension  and 
compression,  and  its  value  is  about  one-half  of  the  ultimate 
shearing  strength.  When  the  shearing  unit-stress  does  not 
exceed  the  elastic  limit,  the  amount  of  slipping  or  detrusion  is 
proportional  to  the  applied  force;  when  it  is  greater,  the  defor- 
mation increases  more  rapidly  than  the  force. 

IT)  /  * 

b 


Fig.  Qc  Fig.  Qd 

Wooden  specimens  for  tensile  tests,  like  that  shown  in  Fig.  6c, 
will  fail  by  shearing  off  the  ends  if  their  length  is  not  sufficiently 
great.  For  example,  let  ab  be  6  inches  and  the  diameter  of  the 
central  part  be  ij  inches.  The  ends  are  gripped  tightly  in  the 
testing  machine  and  the  cross-section  of  the  central  part  thus 
brought  under  tensile  stress.  The  load  required  to  cause  rup- 
ture by  tension  is, 

P=aS=o.7854Xi.52Xioooo  =  i7  700  pounds 

But  the  ends  may  also  shear  off  on  the  surface  of  a  cylinder  hav- 
ing the  length  ab  and  a  diameter  of  ij  inches;  the  load  required 
to  cause  this  rupture  by  shearing  along  the  grain  is, 

P=aS=s.  142X1. 5X6X500=14  100  pounds 

and  hence  the  specimen  will  fail  by  shearing  off  the  ends  before 
the  tensile  strength  of  the  timber  is  reached.  To  prevent  this 
shearing  the  length  ab  must  be  made  about  8  inches. 

When  a  bar  is  subject  either  to  tension  or  to  compression, 


16  ELASTIC  AND  ULTIMATE  STRENGTH  CHAP.  I 

a  shear  occurs  in  any  oblique  section.  Let  Fig.  Qd  represent 
a  bar  of  section  area  a  subject  to  the  tension  P  which  produces 
in  every  normal  section  the  unit-stress  P/a.  Let  mn  be  a  plane 
making  an  angle  6  with  the  axis  of  the  bar  and  cutting  from 
the  bar  a  section  having  the  area  a\.  On  the  left  of  the  plane, 
the  normal  stress  P  may  be  resolved  into  the  components  P\ 
and  P%,  respectively  parallel  and  normal  to  the  plane,  and  the 
same  may  be  done  on  the  right.  Thus  it  is  seen  that  the  effect 
of  the  tensile  force -on  the  plane  mn  is  to  produce  a  tension  P2 
normal  to  it  and  a  shear  PI  along  it,  for  the  two  forces  PI  and 
PI  act  in  opposite  directions  on  opposite  sides  of  the  oblique  sec- 
tion. The  shearing  stress  PI  has  the  value  P  cos#  which  is  dis- 
tributed over  the  area  a\  and  this  area  equals  a/smd.  Hence 
the  shearing  unit-stress  along  the  oblique  section  is, 
Si=Pi/ai  =  (P/a)  sinfl  cos6  =  $(P/a)  $11126 

When  #=o°  or  when  6=  90°,  the  value  of  S\  is  zero,  that  is  there 
is  no  shear  on  a  plane  parallel  with  or  normal  to  the  axis.  For 
all  other  values  of  0,  a  shearing  stress  exists;  and  the  maximum 
value  of  Si  occurs  when  #=45°,  and  then  S\=\P/a.  There- 
fore a  normal  tensile  unit-stress  S  on  a  bar  produces  a  shearing 
unit-stress  JS  along  every  section  inclined  45  degrees  to  the  axis 
of  the  bar.  This  investigation  applies  also  to  compression  and 
it  partially  explains  why  the  rupture  of  compression  specimens 
sometimes  occurs  by  shearing  along  oblique  sections,  as  indicated 
in  Fig.  5c. 

Prob.  Ga.  A  wrought-iron  bolt  ij  inches  in  diameter  has  a  head  ij 
inches  long,  and  a  tension  of  38000  pounds  is  applied  longitudinally 
to  it.  Compute  the  tensile  unit-stress,  and  the  factor  of  safety  against 
tension.  Compute  the  unit-stress  tending  to  shear  off  the  head  of  the 
bolt,  and  the  factor  of  safety  against  shear. 

ART.  7.    WORKING  UNIT-STRESSES 

When  a  bar  of  section  area  a  is  under  axial  stress  caused 
by  a  load  P,  the  unit-stress  S  produced  is  found  by  dividing  P 
by  a.  By  comparing  this  value  of  S  with  the  ultimate  strengths 
and  elastic  limits  given  in  the  preceding  articles,  the  degree  of 
security  of  the  bar  may  be  inferred.  This  process  is  called 


ART.  7  WORKING  UNIT-STRESSES  17 

investigation.  The  student  may  not  at  first  be  able  to  form  a 
good  judgment  with  regard  to  the  degree  of  security,  this  being 
a  matter  which  involves  some  experience,  as  well  as  acquaintance 
with  engineering  precedents  and  practice.  As  his  knowledge 
increases,  however,  his  ability  to  judge  whether  unit-stresses 
are  or  are  not  too  great  will  constantly  improve. 

When  a  bar  is  to  be  designed  to  resist  an  axial  load  P  applied 
at  each  end,  the  unit-stress  S  is  to  be  assumed  in  accordance 
with  the  rules  of  practice,  and  then  the  section  area  a  is  to  be 
computed.  Such  assumed  unit-stresses  are  often  called  'Work- 
ing Unit-Stresses ',  meaning  that  these  are  the  unit-stresses  under 
which  the  material  is  to  act  or  work.  In  selecting  them,  two 
fundamental  rules  are  to  be  kept  in  mind: 

1.  Working  unit-stresses  should  be  considerably  less  than 
the  elastic  limit  of  the  material. 

2.  Working  unit-stresses  should  be  smaller  for'  sudden 
loads  than  for  steady  loads. 

The  reason  for  the  first  requirement  is  given  in  Art.  2.  The 
reason  for  the  second  requirement  is  that  experience  teaches 
that  suddenly  applied  loads  and  shocks  are  more  injurious  and 
produce  larger  deformations  than  steady  loads.  Thus,  a  bridge 
subject  to  the  traffic  of  heavy  trains  must  be  designed  with  lower 
unit-stresses  than  a  roof  where  the  variable  load  consists  only 
of  snow  and  wind.  In  buildings  the  stresses  are  mostly  steady, 
in  bridges  they  are  variable,  and  in  machines  the  stresses  are 
often  produced  with  shock. 

It  will  be  best  for  the  student  to  begin  to  form  his  engineer- 
iftg  judgment  by  fixing  in  mind  the  following  average  values 
of  the  factors  of  safety  to  be  used  for  different  materials  under 
different  circumstances : 

Material  Steady  Stress    Variable  Stress       Shocks 

Brick  and  Stone  15  25  40 

Timber  8  10  15 

Cast  Iron  6  10  20 

Wrought  Iron  4  6  10 

Structural  Steel  4  6  10 

Strong  Steel  5  8  15 


18  ELASTIC  AND  ULTIMATE  STRENGTH  CHAP,  i 

The  working  unit-stress  will  then  be  found  for  any  special  case 
by  dividing  the  ultimate  strengths  by  these  factors  of  safety. 
For  instance,  a  short  timber  strut  in  a  bridge  should  have  a  work- 
ing unit-stress  of  about  8000/10  =  800  pounds  per  square  inch. 

It  is  usually  the  case  that  a  designer  works  under  specifica- 
tions in  which  the  unit-stresses  to  be  used  are  definitely  stated. 
The  writer  of  the  specifications  must  necessarily  be  an  engineer 
of  much  experience  and  with  a  thorough  knowledge  of  the  best 
practice.  The  particular  qualities  of  timber  or  steel  to  be  used 
will  influence  his  selection  of  working  unit-stresses,  and  in  fact 
different  members  of  a  bridge  truss  are  often  designed  with 
different  unit-stresses.  The  two  fundamental  rules  above  given 
are,  however,  the  guiding  ones  in  all  cases.  In  Table  5,  at  the 
end  of  this  volume,  will  be  found  the  working  unit-stresses  which 
are  specified  in  the  building  code  of  the  City  of  New  York.  These 
apply  in  the  design  of  buildings,  and  the  unit-stresses  to  be  used 
in  the  design  of  bridges  are  lower  than  those  in  the  table,  while 
those  to  be  used  in  the  design  of  machines  should  be  lower  still. 

The  two  fundamental  principles  of  engineering  design  are 
stability  and  economy,  or  in  other  words : 

1.  A  structure  must  safely  withstand  all  the  stresses  which 
may  be  caused  by  loads  that  can  come  upon  it. 

2.  A  structure  should  be  designed  so  that  it  may  be  built 
and  maintained  at  the  lowest  possible  cost. 

The  second  of  these  fundamental  principles  requires  that  all 
parts  of  the  structure  should  be  of  equal  strength,  like  the  cele- 
brated 'one-hoss  shay'  of  the  poet.  For,  if  one  part  is  stronger 
than  another,  it  has  an  excess  of  material  which  might  have 
been  saved.  Of  course  this  rule  is  to  be  violated  when  the  cost 
of  the  labor  required  to  save  the  material  is  greater  than  that  of 
the  material  itself.  Thus  it  often  happens  that  some  parts  of 
a  structure  have  higher  factors  of  safety  than  others,  but  the 
lowest  factors  should  not  be  less  than  those  which  good  engineer- 
ing practice  requires. 

The  factors  of  safety  stated  above  may  be  supposed  to  be 
so  arranged  that,  if  different  materials  are  united,  the  stabilitv 


ART.  8  COMPUTATIONS  AND  EQUATIONS  19 

of  all  parts  of  the  structure  will  be  the  same,  so  that  if  rupture 
occurs,  everything  would  break  at  once.  Or,  in  other  words, 
timber  with  a  factor  of  safety  8  has  about  the  same  reliability 
as  steel  with  a  factor  of  4  or  stone  with  a  factor  of  15,  provided 
the  stresses  are  due  to  steady  loads. 

As  an  example  of  design,  let  it  be  required  to  determine  the 
size  of  a  short  piston-rod  made  of  hard  steel  having  an  ultimate 
tensile  strength  of  75000  pounds  per  square  inch,  the  piston 
being  20  inches  in  diameter  and  the  steam  pressure  upon  it  being 
80  pounds  per  square  inch.  This  being  a  rod  subject  to  rapidly 
alternating  stresses  of  tension  and  compression  and  perhaps  to 
shocks,  the  factor  of  safety  should  be  taken  at  15,  which  gives  a 
working  unit-stress  of  75  000/15  =  5  ooo  pounds  per  square  inch. 
The  total  tension  being  0.7854X202x80  =  25  100  pounds,  the 
area  of  the  cross-section  should  be  25100/5000  =  5.03  square 
inches  and  this  will  be  furnished  by  a  rod  2.53  inches  in  diam- 
eter, so  that  a  diameter  of  2\  inches  will  probably  be  sufficient 
to  give  proper  security  against  tension.  Since  the  compressive 
strength  of  hard  steel  is  higher  than  the  tensile  strength,  and 
since  the  rod  is  short,  this  diameter  also  gives  proper  security 
against  compression. 

Prob.  7a.  A  rod  of  structural  steel  is  to  be  used  under  a  tension  of 
87  ooo  pounds.  Compute  its  diameter  when  it  is  to  be  used  in  a 
building,  and  also  when  it  is  to  be  used  in  a  bridge. 

Prob.  7b.  In  Fig.  Qb  each  of  the  loads  is  4  700  pounds  and  the 
wooden  beam  is  2X3  inches  in  cross-section.  Compute  the  factor  of 
safety  against  shearing.  Is  this  factor  sufficiently  high  for  steady 
loads  ? 

ART.  8.    COMPUTATIONS  AND  EQUATIONS 

The  numerical  computations  required  in  the  mechanics  of 
materials  should  be  performed  so  that  the  precision  of  the  results 
fairly  corresponds  with  the  precision  of  the  data.  The  values  of 
the  elastic  limits  and  ultimate  strengths  given  in  the  preceding 
tables  are  indefinite  in  the  second  significant  figure  and  hence 
computed  areas  should  not  be  carried  further  than  three  sig- 


20  ELASTIC  AND  ULTIMATE  STRENGTH  CHAP.  I 

nificant  figures  the  last  of  which  has  no  precision.  For  instance 
in  the  area  of  5.03  square  inches  computed  at  the  end  of  the 
last  article  the  last  figure  is  of  no  importance,  since  the  work- 
ing unit-stress  of  5  ooo  pounds  per  square  inch  is  derived  from 
rough  data.  Numerical  computations,  therefore,  should  include 
only  three  significant  figures  as  a  general  rule. 

At  the  end  of  this  volume  are  given  tables  of  four-place  loga- 
rithms, squares  of  numbers,  and  areas  of  circles  which  are  of 
value  in  abridging  computations.  The  table  of  squares  may  be 
used  to  find^quare  roots,  and  the  table  of  circles  to  find  diam- 
eters from  given  areas.  For  example,  a  circle  having  an  area 
of  0.82  square  inches  has  a  diameter  of  1.02  inches;  a  circle 
having  an  area  of  8.2  square  inches  has  a  diameter  of  3.23  inches. 

As  this  book  is  mainly  intended  for  the  use  of  students  in 
engineering  colleges,  a  word  of  advice  directed  especially  to 
them  may  not  be  inappropriate.  It  will  be  necessary  for  stu- 
dents, in  order  to  gain  a  clear  idea  of  the  mechanics  of  materials, 
or  of  any  engineering  subject,  to  solve  many  numerical  problems, 
and  in  this  work  a  neat  and  systematic  method  should  be  followed. 
The  practice  of  making  computations  on  any  loose  scraps  of 
paper  that  may  happen  to  be  at  hand  should  be  discontinued 
by  every  student  who  has  followed  it,  and  he  should  hereafter 
solve  his  problems  in  a  special  book  provided  for  that  purpose, 
accompanying  them  by  explanatory  remarks.  Such  a  note- 
book, written  in  ink,  and  containing  the  solutions  of  the  prob- 
lems given  in  these  pages  will  prove  of  great  value  to  every  stu- 
dent. Before  beginning  the  solution  of  a  problem,  a  diagram 
should  be  drawn  whenever  possible,  for  a  diagram  helps  the 
student  to  understand  the  problem,  and  a  problem  thoroughly 
understood  is  half  solved.  Before  beginning  the  numerical  work, 
it  is  also  well  to  make  a  mental  estimate  of  the  answer  and  record 
the  same,  comparing  it  later  with  the  result  of  the  solution,  since 
in  this  manner  the  engineering  judgment  of  the  student  will  be 
developed. 

In  continental  Europe  the  metric  system  is  universally  employed 
in  the  mechanics  of  materials,  the  unit  of  force  or  stress  being 


A.RT.  8       COMPUTATIONS  AND  EQUATIONS         21 

the  kilogram,  that  of  length  being  the  meter  or  centimeter,  and 
that  of  area  being  the  square  centimeter.  Computations  in  the 
metric  system  are  much  simpler  than  those  in  the  English  sys- 
tem, and  it  is  to  be  hoped  that  the  time  is  not  far  distant  when 
it  will  come  into  general  use.  As  a  slight  contribution  to  this 
end,  the  average  constants  relating  to  the  strength  of  materials 
are  given  in  metric  units  in  Tables  1-4  at  the  end  of  this  volume, 
and  a  few  problems  using  such  units  will  be  occasionally  pro- 
posed. In  solving  these  problems  the  student  should  think  in 
the  metric  system  and  make  no  transformations. of  the  data  or 
the  results  into  the  English  system.  The  table  of  areas  of  circles 
at  the  end  of  this  volume  is  applicable  to  all  systems  of  measures. 

In  the  English  system  of  measures,  as  generally  used  in  this 
volume,  the  unit  of  force  is  the  pound,  the  unit  of  length  is  the 
inch,  and  the  unit  of  area  is  the  square  inch.  Lengths  of  bars 
and  beams  are  sometimes  stated  in  feet,  but  these  should  be 
reduced  to  inches  when  they  are  to  be  used  in  formulas  in  order 
that  they  may  be  consistent  with  the  other  data. 

In  this  volume  Greek  letters  are  used  only  for  signs  of  opera- 
tion, for  abstract  numbers,  and  for  angles.     The  letter  d  is  em- 
ployed as  the  symbol  of  differentiation  and  I  as  the  symbol  of 
summation;   the  Greek  names  of  these  two  letters  should  not  be 
used,  but  they  may  be  called  'differential'  and  'algebraic  sum'. 
The  following  are  the  names  of  other  Greek  letters: 
a  Alpha,        /?  Beta,          e  Epsilon,       i)  Eta, 
6  Theta,        K  Kappa,      A  Lambda,     /*  Mu, 
v  Nu,  TT  Pi,  p  Rho,  o  Sigma, 

r  Tau,  ^  Phi,  ([>  Psi,  CD  Omega. 

In  every  algebraic  equation  it  is  necessary  that  all  of  the 
terms  should  be  of  the  same  dimension,  for  it  is  impossible  to 
add  together  quantities  of  different  kinds.  This  principle  will 
be  of  great  assistance  to  the  student  in  checking  the  correctness 
of  algebraic  work.  For  example,  let  a  and  b  represent  areas 
and  /  a  length;  then  such  an  equation  as  al  —  l2=b  is  impossible, 
because  al  is  a  volume  while  I2  and  b  are  areas.  Again,  let  S 
represent  pounds  per  square  inch,  P  pounds,  /  inches,  and  a 


22  ELASTIC  AND  ULTIMATE  STRENGTH  CHAP.  I 

square  inches;  then  the  equations  S=P/a  and  P=aS  are  cor- 
rect with  respect  to  dimensions,  but  the  equation  S2l2/P=a  is 
impossible  because  the  first  member  represents  pounds  per 
square  inch  while  the  second  is  an  area.  The  equation 
Si  =\(P/a)  sin2#,  deduced  in  Art.  6,  is  correct  in  dimensions, 
for  both  Si  and  P/a  represent  pounds  per  square  inch  while 
sin20  is  an  abstract  number. 

Prob.  8a.  A  bar  of  wrought  iron  is  3.25  centimeters  in  diameter. 
What  load  in  kilograms  will  cause  it  to  rupture  by  tension?  What 
load  will  stress  it  up  to  one-half  of  the  elastic  limit  ? 

Prob.  8b.  A  round  bar  of  structural  steel  is  under  a  tension  of  7  ooo 
kilos.  What  should  be  its  diameter  in  centimeters  in  order  that  the 
factor  of  safety  may  be  6  ? 

Prob.  8c.  Let  K  represent  work,  P  pounds,  S  and  E  pounds  per 
square  inch,  /  inches,  and  a  square  inches;  determine  whether  or 
not  the  formula  K  =  %(S2/E)al  is  dimensionally  correct.  Also  show 
whether  the  equation  x5+pxP+qx-t-pq  =  o  is  or  is  not  correct. 


ART.  9  MODULUS   OF   ELASTICITY  23 


CHAPTER  II 
ELASTIC  AND  ULTIMATE  DEFORMATION 

ART.  9.     MODULUS  OF  ELASTICITY 

The  term  'Elastic  Deformation'  is  used  to  designate  that 
change  of  shape  of  a  body  which  accompanies  stresses  that  do 
not  surpass  the  elastic  limit  of  the  material.  In  tension  the 
principal  deformation  is  the  elongation  of  the  bar,  in  compres- 
sion it  is  the  shortening.  The  fact  that  these  deformations  are 
proportional  to  the  stresses  (Art.  2)  enables  rules  to  be  estab- 
lished whereby  the  change  of  length  of  a  bar  can  be  computed, 
provided  the  elastic  limit  be  not  exceeded. 

Let  P  be  an  axial  tensile  load  applied  to  a  bar  which  has  a 
cross-section  of  area  a;  then  the  tensile  unit-stress  S  is  P/a 
(Art.  1).  Let  /  be  the  length  of  the  bar  and  e  the  total  elonga- 
tion which  occurs  under  the  stress ;  then  the  unit-elongation  is  e/l, 
and  this  will  be  designated  by  e.  Similarly  let  P  be  a  compres- 
sive  load  which  produces  the  shortening  e,  the  compressive  unit 
stress  is  P/a,  and  the  unit-shortening  is  e/l.  Unit-stress  is  the 
stress  on  a  unit  of  area,  the  total  stress  being  regarded  as  uni- 
formly distributed  over  the  section  area  of  the  bar.  Unit-deforma- 
tion is  the  change  in  length  of  a  linear  unit  of  the  bar,  the  total 
deformation  being  regarded  as  uniformly  distributed  over  the  total 
length.  Accordingly  for  any  bar  in  tension  or  compression, 

Unit-stress  S= P/a  Unit-deformation  s  =  e/l 

and  these  are  applicable  whether  the  elastic  limit  be  exceeded 
or  not,  as  illustrated  in  Art.  4. 

When  the  elastic  limit  is  not  exceeded  by  the  unit-stress,  the 
unit-deformation  e/l  is  proportional  to  the  unit-stress  P/a,  and 
hence  the  ratio  of  the  latter  to  the  former  is  a  constant  for  each 
kind  of  material.  The  term  ' Modulus  of  Elasticity'  is  used  for 
this  constant,  and  it  may  be  denned  as  the  ratio  of  the  unit-stress 


24  ELASTIC  AND  ULTIMATE  DEFORMATION         CHAP.  II 

to  the  unit-deformation.  The  letter  E  is  used  for  the  modulus 
of  elasticity  and  hence,  from  the  definition,  its  value  for  tension 
or  compression  is, 

77    s  v    p/a 

E  =  —  or  E  =  ~TT  (9 

£  e/l 

Since  e  and  /  are  linear  quantities,  e  is  an  abstract  number,  and 
therefore  E  is  expressed  in  the  same  unit  as  S,  that  is  in  pounds 
per  square  inch  or  in  kilograms  per  square  centimeter. 

The  above  equations  show  that  E  equals  P  when  a  is  unity 
and  e  is  equal  to  /;  that  is,  the  modulus  of  elasticity  is  the  force 
which  will  elongate  a  bar  of  one  square  unit  in  cross-section  to 
double  its  original  length,  provided  that  this  can  be  done  with- 
out exceeding  the  elastic  limit  of  the  material.  There  is  prob- 
ably no  material,  except  india-rubber,  for  which  so  great  an 
elastic  elongation  is  possible. 

The  modulus  of  elasticity  E  is  called  by  some  writers  the 
'coefficient  of  elasticity',  but  the  former  term  is  now  of  more 
general  use  in  the  United  States  of  America;  in  books  on  physics 
it  is  often  called  'Young's  modulus'.  The  student  should  care- 
fully note  that  the  above  formulas  for  E  apply  only  when  the 
unit-stress  S  or  P/a  is  less  than  the  elastic  limit  of  the  material. 
When  modulus  of  elasticity  is  mentioned  without  qualification 
it  is  always  understood  to  refer  to  tension  or  compression  and 
not  to  shear  (Art.  15). 

When  gradually  increasing  loads  are  applied  to  a  bar  and 
the  values  of  e  are  measured  for  different  values  of  P,  the  simul- 
taneous quantities  P/a  and  e/l  are  known,  and  their  ratio  gives 
the  value  of  E.  The  following  are  approximate  average  values 
of  E  for  the  different  materials  used  in  engineering  construction, 
which  have  been  derived  from  the  records  of  numerous  tests : 

Material  Modulus  of  Elasticity 

Brick  2  ooo  ooo  pounds  per  square  inch 

Stone  6  ooo  ooo  pounds  per  square  inch 

Timber  i  500  ooo  pounds  per  square  inch 

Cast  Iron  15  ooo  ooo  pounds  per  square  inch 

Wrought  Iron  25  ooo  ooo  pounds  per  square  inch 

Steel  30  ooo  ooo  pounds  per  square  inch 


ART.  10  ELASTIC  CHANGE  IN  LENGTH  25 

The  values  here  given  for  brick  and  stone  apply  only  to  com- 
pression and  little  is  known  regarding  the  elastic  properties  of 
these  materials  under  tension;  those  given  for  the  other  mate- 
rials apply  to  both  tension  and  compression.  It  is  seen  that 
the  modulus  of  elasticity  for  timber  is  one-tenth  of  that  for  cast 
iron,  that  for  cast  iron  being  one-half  of  that  for  steel.  Table  2, 
at  the  end  of  this  book,  gives  these  constants  in  metric  measures. 

The  modulus  of  elasticity  is  a  measure  of  the  stiffness  of 
the  material,  that  is,  of  its  ability  to  resist  change  of  shape  under 
unit-stresses  not  higher  than  the  elastic  limit.  For,  the  unit- 
deformation  £  may  be  expressed  by  S/E,  and  hence  e  is  the 
least  for  a  given  S  when  E  is  the  greatest.  The  above  values 
of  E  show  that  the  elastic  change  in  length  of  a  steel  bar  is  one- 
twentieth  of  that  of  a  timber  bar  and  one-half  of  that  of  a  cast- 
iron  bar,  the  applied  tension  being  the  same  in  the  three  cases 
and  the  sizes  and  lengths  of  the  bars  being  equal. 

Prob.  9a.  Compute  the  modulus  of  elasticity  of  a  bar,  ij  inches  in 
diameter  and  16  feet  long,  which  elongates  0.125  inches  under  a  ten- 
sion of  21  ooo  pounds. 

Prob.  96.  A  wooden  specimen  i  inch  in  diameter  and  18  inches 
Jong,  elongates  0.013  inches  when  the  tension  is  increased  from  800  to 
\  600  pounds,  and  0.24  inches  when  the  tension  is  increased  from 
i  600  to  6  ooo  pounds.  Compute  the  modulus  of  elasticity. 

ART.  10.    ELASTIC  CHANGE  IN  LENGTH 

The  change  in  length  of  a  bar  under  an  axial  stress  which 
does  not  exceed  the  elastic  limit  of  the  material  is  readily  com- 
puted when  the  modulus  of  elasticity  of  the  material  is  known. 
From  the  definition  of  that  modulus  in  Art.  9,  it  is  seen  that  the 
unit-deformation  in  length  is  e  =  S/Et  and  that  the  total  deforma- 
tion in  length  is 

-i'    -    -¥<-5 

in  which  e  is  the  change  in  the  length  /,  and  P/a  or  S  is  the  unit- 
stress,  while  E  is  the  modulus  of  elasticity.  If  e  is  to  be  found 
in  inches,  then  /  must  be  in  inches;  S  and  E  are  in  pounds  per 


26  ELASTIC  AND  ULTIMATE  DEFORMATION        CHAP.  II 

square  inch  and  hence  their  ratio  is  an  abstract  number.  When 
S  is  a  tensile  unit-stress,  e  is  the  elongation  of  the  bar;  when 
S  is  a  compressive  unit-stress,  e  is  the  shortening  of  the  bar. 

These  formulas  may  be  used  for  computing  the  change  of 
length  which  occurs  at  the  elastic  limit,  or  under  any  value  of  S 
less  than  the  elastic  limit.  Using  the  mean  values  of  the  elastic 
limits  in  Art.  2  and  those  of  E  in  Art.  9,  the  following  values  of 
e/l  are  found  for  tension : 

For  timber,  e/l=  1/500  -=0.0020 

For  cast  iron,  e/l  =  1/2  500  =  0.0004 

For  wrought  iron,  e/l= 1/1000=0.0010 

For  structural  steel,  e/l=  1/850  =0.0012 

These  quantities  show  that,  when  bars  of  the  same  length 
are  stressed  up  to  their  elastic  limits,  the  elongation  of  the  cast- 
iron  bar  is  the  least,  that  of  wrought  iron  next,  and  that  of  timber 
the  greatest.  The  ultimate  elongations,  namely  those  which 
obtain  at  rupture,  follow  a  different  order  and  are  very  much 
greater  than  those  which  occur  at  the  elastic  limit. 

As  an  example  for  stresses  within  the  elastic  limit,  let  it  be 
required  to  find  the  elongation  of  a  bar  of  steel,  i  J  X 1 2  inches 
in  section  area  and  23  feet  long,  when  under  a  tension  of  270  ooo 
pounds.  The  unit-stress  under  this  tension  is  5  =  270000/18  = 
15  ooo  pounds  per  square  inch,  and  as  this  is  less  than  the  elastic 
limit  the  formula  applies,  and  0=15  000X276/30000000  =  0.138 
inches.  This  would  also  be  the  amount  of  shortening  of  the 
bar  under  a  compression  of  270000  pounds,  provided  that  no 
lateral  bending  occurred,  but  it  will  be  seen  later  that  a  bar  of 
this  size  and  length  is  unable  to  withstand  a  unit-stress  as  high 
as  15  ooo  pounds  per  square  inch  on  account  of  the  sidewise 
flexure.  The  above  formulas  can  in  general  be  used  only  for 
finding  the  shortening  of  a  bar  in  compression  when  its  length 
is  less  than  twenty  times  its  thickness. 

From  the  formula  e/l  =  (P/a)/E,  which  applies  to  all  elastic 
changes  of  length,  any  one  of  the  five  quantities  may  be  com- 
puted when  the  other  four  are  given.  For  example,  the  unit- 
elongation  of  timber  when  it  is  stressed  up  to  one-half  of  its 


ART.  11  ELASTIC  LIMIT  AND  YIELD  POINT  27 

elastic  limit  is  found  by  e//=  1500/1  500000  =  1/1000,  so  that 
if  a  stick  of  timber  elongates  0.48  inches,  its  length  is 
1  =  0.48X1  000  =  480  inches  =  40  feet.  Should  it  be  required  to 
find  the  elongation  of  a  bar  of  wrought  iron,  if  inches  in  diam- 
eter and  1 6  feet  long,  under  a  tension  of  51  ooo  pounds,  the 
section  area  is  1.485  square  inches,  and  the  unit-stress 
5  =  51000/1.485=34300  pounds  per  square  inch;  as  this  is 
greater  than  the  elastic  limit  of  the  material,  the  formula  has  no 
validity  and  it  is  impossible  to  compute  by  it  the  value  of  the 
elongation. 

Prob.  10a.  Compute  the  elongations  of  a  wrought-iron  bar,  if 
inches  in  diameter  and  16  feet  long,  under  tensions  of  10  ooo,  20  ooo, 
and  30  ooo  pounds. 

Prob.  lOb.  Compute  the  tensile  force  required  to  stretch  a  bar  of 
structural  steel,  ifX9f  inches  in  section  area  and  23  feet  3^  inches 
long,  so  that  its  length  may  be  increased  to  23  feet  5^  inches. 


ART.  11.    ELASTIC  LIMIT  AND  YIELD  POINT 

In  Art.  2  the  elastic  limit  was  denned  as  that  unit-stress  within 
which  the  deformation  of  a  bar  is  proportional  to  the  stress  and 
beyond  which  the  deformation  increases  in  a  more  rapid  ratio 
than  the  stress.  The  diagram  in  Art.  4  illustrates  these  experi- 
mental facts  for  tension,  the  relation  between  unit-stress  and 
unit-elongation  being  shown  by  a  straight  line  until  the  elastic 
limit  is  reached  and  afterwards  by  a  curve.  The  point  where 
the  straight  line  is  tangent  to  the  curve  indicates  the  elastic  limit 
of  the  material. 

In  Fig.  \\a  there  is  given,  on  a  larger  horizontal  scale,  a  part 
of  the  tension  diagram  of  Fig.  4.  For  each  material  the  point 
of  elastic  limit  is  marked  by  a  dash  normal  to  the  curve.  For 
any  unit-stress  S~less  than  the  elastic  limit,  the  ratio  of  S  to  the 
unit-deformation  e  is  a  constant,  or  S/e  =  E,  and  E  is  the  modu- 
lus of  elasticity  of  the  material.  The  greater  the  inclination  of 
the  straight  line  to  the  horizontal,  the  greater  is  the  value  of  E. 
Since  £  and  S  are  variable  rectangular  coordinates,  the  equation 


28 


ELASTIC  AND  ULTIMATE  DEFORMATION        CHAP.  II 


S  =  Ee  is  that  of  a  straight  line  and  E  is  the  tangent  of  the  angle 
vvhich  this  line  makes  with  the  horizontal. 


60000 


o.ooi  0.002 

Unit-Elongation  = 


0.003 


Fig.  lla 

Another  definition  of  the  elastic  limit  is  that  it  is  the  unit- 
stress  within  which  the  bar  returns  to  its  original  length,  when 
the  load  is  removed,  and  beyond  which  it  does  not  fully  return, 
a  part  of  the  deformation  being  permanent.  The  term  '  limit 
of  elasticity'  is  sometimes  used  instead  of  elastic  limit  when 
considering  the  subject  from  this  point  of  view.  Fig.  lib  shows 
a  part  of  the  tensile  diagram  for  wrought  iron  in  which  AB  repre- 
sents a  bar  of  unit  length  which  is  stretched  to  the  lengths  AB?, 
AC' ,  and  AD'  under  stresses  of  25  ooo,  30  ooo,  and  34  ooo  pounds 
per  square  inch.  If  the  tension  is  removed  when  the  stretch  is 
less  than  BB'  the  bar  springs  back  to  its  original  length  AB\ 
here  B'b  is  the  elastic  limit  of  the  material.  If  the  tension  is 
removed  when  the  bar  has  the  stretch  BD' ',  the  bar  springs  back 
to  D  so  that  its  length  is  AD  and  it  has  the  permanent  set  BD. 
Similarly  for  any  length  AC  the  bar  under  tension  has  the  stretch 
BCf  and,  on  the  removal  of  the  tension,  it  springs  back  to  the 
length  AC  and  has  the  permanent  set  BC.  It  has  been  found 
by  many  experiments  that  the  lines  Cc  and  Dd  are  nearly  straight 
and  closely  parallel  to  the  line  Bb. 

An  inspection  of  Figs,  lla  and  116  shows  that  it  is  more  diffi- 
cult to  locate  the  point  of  tangency  on  the  lines  for  timber  and 
cast  iron  than  on  those  for  wrought  iron  and  structural  steel. 


ART.  11 


ELASTIC  LIMIT  AN»  YIELD  POINT 


29 


In  other  words  the  elastic  limits  in  tension  for  timber  and  cast 
iron  are  not  well  defined.  On  the  compression  diagram  it  is 
also  seen  that  the  elastic  limits  for  wrought  iron  and  structural 
steel  are  not  so  sharply  defined  as  those  for  tension,  although 
they  are  more  determinable  than  for  the  other  materials. 

The  'Yield  Point'  is  defined  to  be  that  unit-stress  at  which 
the  deformation  increases  without  any  increase  in  applied  load 
or  in  internal  stress.  In  Fig.  lla  it  is  seen  that  the  lines  for 
wrought  iron  and  structural  steel  curve  beyond  the  elastic  limit 
for  a  short  distance  and  then  become  horizontal.  In  Fig.  116 
the  point  where  the  curve  becomes  horizontal  is  marked  by  the 
letter  b',  and  the  unit-stress  corresponding  to  this  is  called  the 
yield  point.  Only  ductile  materials  like  wrought  iron  and  struc- 
tural steel  have  yield  points  and  these  generally  belong  to  ten- 
sion and  are  usually  only  observed  in  testing  machines  where 
the  tension  is  slowly  applied.  Beyond  the  yield  point  the  curve 
continues  horizontal  for  a  short  distance  and  then  gradually  rises. 

40000 


Fig.  lib 

In  commercial  testing  the  yield  point  is  sometimes  called  the 
elastic  limit  because  the  former  is  more  easily  ascertained  than 
the  latter  (Art.  25).  This  is  an  improper  use  of  the  term  elastic 
limit  which  should  be  avoided.  For  structural  steel  the  yield 
point  is  often  higher  than  the  elastic  limit  by  from  3  ooo  to  6  ooo 
pounds  per  square  inch,  and  hence  it  is  important  that  the  dis^- 
tinction  should  be  carefully  observed. 

Prob.  11.  Bars  of  cast  iron,  wrought  iron,  and  ste:l,  i  square  inch  in 
section  area  and  100  inches  long,  are  subject  to  a  tension  of  15000 
pounds.  Determine  the  elongation  of  each  bar  from  Fig.  lla.  Also 
compute  the  elongations,  as  far  as  possible,  from  the  rules  of  Art.  10. 


30  ELASTIC  AND  ULTIMATE  DEFORMATION        CHAP.  II 


ART.  12.    ULTIMATE  DEFORMATIONS 

Mean  values  of  the  ultimate  unit-elongations  for  different 
materials  are  given  in  the  table  of  Art.  4,  and  the  stress  diagrams 
in  Fig.  4  show  that  these  are  large  compared  with  that  which 
occurs  at  the  elastic  limit.  Timber  and  cast-iron  bars  have 
ultimate  elongations  about  8  times  as  great  as  the  elastic  elonga- 
tions, while  steel  has  ultimate  elongations  from  100  to  300  times 
as  great  as  the  elastic  ones.  There  is  no  method  by  which  the 
ultimate  elongation  of  a  bar  can  be  computed,  but  all  knowledge 
concerning  it  must  be  derived  from  the  records  of  tests. 

The  ultimate  elongation  of  a  bar  is  determined  by  making 
two  marks  upon  it  before  it  is  subjected  to  tension  and  measuring 
the  distance  between  them  before  and  after  the  test.  The  differ- 
ence of  these  lengths  divided  by  the  original  length  gives  the  ulti- 
mate elongation  per  unit  of  length,  that  is,  the  unit-elongation. 
For  example,  if  the  distance  between  the  two  marks  is  2.01  inches 
and  if  this  becomes  2.57  inches  after  the  rupture,  then  the  total 
elongation  is  2.57  —  2.01=0.56  inches,  and  the  ultimate  unit- 
elongation  is  0.56/2.01  =0.280  or  28.0  percent. 

In  Fig.  185&  are  shown  three  steel  specimens,  that  at  the  top 
being  one  which  has  not  been  tested.  Its  total  length  is  5^  inches 
and  about  i\  inches  of  the  central  part  is  one-half  an  inch  in 
diameter.  The  marks  are  placed  on  this  cylinder  about  2  inches 
apart,  the  exact  distance  being  measured  to  the  nearest  hundredth 
of  an  inch.  The  two  other  specimens  have  been  ruptured  by 
tension  applied  through  the  screw  ends  by  the  testing  machine, 
and  the  respective  elongations  were  found  to  be  3.8  and  22.5 
percent.  This  great  difference  in  the  ultimate  elongation  of  steel 
may  be  due  to  differences  in  chemical  composition  and  method 
of  manufacture,  but  in  this  case  it  was  largely  due  to  a  flaw  in 
the  ruptured  section  of  the  middle  specimen.  The  loads  that 
ruptured  these  two  specimens  were  18  600  and  22  ooo  pounds 
respectively,  so  that  the  ultimate  strengths  were  about  83  ooo 
and  1 10  ooo  pounds  per  square  inch. 


AKT.  12  ULTIMATE  DEFORMATIONS  31 

An  elongation  of  a  bar  is  always  accompanied  by  a  reduc- 
tion in  the  area  of  its  cross-section.  The  greater  the  ultimate 
elongation  the  less  is  the  area  of  the  ruptured  section.  For  duc- 
tile materials,  like  wrought  iron  and  some  kinds  of  steel,  there 
is  observed  a  very  marked  change,  as  the  ultimate  strength  is 
approached,  on  both  sides  of  the  section  where  rupture  is  about 
to  occur.  In  Fig.  185&  this  is  scarcely  observable  in  the  middle 
specimen,  but  it  is  plainly  seen  in  the  lower  one. 

The  term  'Reduction  of  Area'  refers  to  a  ruptured  specimen 
and  means  the  diminution  in  section  area  per  unit  of  original 
area.  Thus  for  the  third  specimen,  the  original  section  area 
was  0.1995  square  inches,  the  area  of  the  ruptured  section  was 
0.1064  square  inches,  and  hence  the  reduction  of  area  is 
0.0931/0.1995=0.467,  or  46.7  percent.  Instead  of  actually  com- 
puting the  area,  the  squares  of  the  diameters  may  be  used;  thus, 
the  original  diameter  was  0.504  inches,  the  diameter  of  the  rup- 
tured section  was  0.368  inches,  and  the  squares  of  these  are 
0.2540  and  0.1354;  the  reduction  of  area  is  then  0.1186/0.2540  = 
46.7  percent.  Reduction  of  area,  or  contraction  of  area  as  it  is 
often  called,  is  an  index  of  the  ductility  of  the  material,  and  it 
is  generally  regarded  as  a  more  reliable  index  than  elongation, 
because  the  ultimate  unit- elongation  is  subject  to  variation  with 
the  ratio  of  the  length  of  the  specimen  to  its  diameter,  whereas 
the  reduction  of  area  is  more  constant.  In  Art.  169  further 
remarks  regarding  ultimate  elongation  will  be  found. 

Under  compression,  a  cube  or  a  prism  decreases  in  length 
and  the  area  of  its  cross-section  increases  with  the  amount  of 
shortening.  The  ultimate  shortening  is,  however,  rarely  deter- 
mined, and  in  most  cases  it  is  much  less  than  the  ultimate  elon- 
gation in  tension.  The  rupture  of  the  compression  specimen, 
having  a  length  of  less  than  ten  times  its  least  lateral  dimen- 
sion, occurs  usually  by  an  oblique  shearing  which  is  illustrated 
in  Fig.  5c  and  which  will  be  discussed  later  (Art.  147). 

When  a  bar  is  under  tension  exceeding  its  elastic  limit,  and 
this  is  released  by  the  removal  of  the  load,  the  length  of  the  bar 
is  greater  than  before,  as  shown  in  the  last  article.  It  is  impos- 


32  ELASTIC  AND  ULTIMATE  DEFORMATION         CHAP,  n 

sible  to  compute  this  new  length,  but  the  amount  of  change 
after  the  removal  of  the  tension  can  be  ascertained  when  the 
modulus  E  is  known.  In  Fig.  lib  let  the  unit-stress  D'd  be 
represented  by  S\  then,  since  dD  is  parallel  to  bB,  the  tangent 
of  the  angle  dDD'  is  the  modulus  E,  and  hence  on  the  removal 
of  the  tension  the  change  in  unit-elongation  is  DD'=S/E. 
Thus,  if  a  bar  of  structural  steel  20  feet  long  is  stressed  in 
tension  up  to  45  ooo  pounds  per  square  inch,  it  will  shorten, 
when  the  stress  is  released,  the  amount  240  X  45  000/30  ooo  ooo  = 
0.36  inches. 

Although  the  area  of  the  cross-section  decreases  under  ten- 
sion and  increases  under  compression,  the  unit-stress  S  is  always 
obtained  by  dividing  the  load  P  by  the  original  area  a.  This  is 
perhaps  not  strictly  correct,  but  it  is  the  custom  in  practice,  and 
all  stress  diagrams  are  constructed  by  using  unit-stresses  com- 
puted in  this  manner. 

Prob.  12.  Test  specimen  No.  7478  of  the  Watertown  Arsenal  was 
0.798  inches  in  diameter  and  6  inches  long  between  marks.  After  rup- 
ture the  distance  between  marks  was  7.44  inches,  and  the  diameter  of 
the  ruptured  section  was  0.64  inches.  Compute  the  percentages  of 
ultimate  elongation  and  reduction  of  area. 

ART.  13.     CHANGES  IN  SECTION  AND  VOLUME 

When  the  unit-stress  does  not  exceed  the  elastic  limit,  the 
changes  in  the  area  of  the  cross-section  and  in  the  volume 
of  a  body  under  stress  are  very  small,  but  it  is  possible  to  com- 
pute them  approximately  as  shown  below.  When  the  unit-stress 
exceeds  the  elastic  limit,  there  is  no  method  by  which  changes 
in  section  and  volume  can  be  computed. 

Many  measurements  of  the  lateral  dimensions  of  bars  under 
normal  stress  have  proved  that  their  elastic  change  is  propor- 
tional to  the  linear  unit-deformation.  Thus,  let  a  bar  of  length 
I  and  diameter  d  have  the  unit-elongation  e ;  its  increase  in  length 
is  el  and  its  decrease  in  diameter  is  Aed,  where  ^  is  a  number 
which  has  been  found  to  be  about  i  for  cast  iron  and  about  J 
for  steel.  For  example,  the  value  of  e  for  wrought  iron  stressed 


ART.  is  CHANGES  IN  SECTION  AND  VOLUME  33 

to  its  elastic  limit  is  about  o.ooi  (Art.  10),  so  that  the  change  in 
length  of  the  bar  then  is  o.ooi/,  while  the  change  in  its  diameter 
is  JX o.ooi d. 

Let  h  be  any  lateral  dimension  of  length  /,  and  let  h'  and  /' 
be  the  lateral  dimension  and  length  under  a. stress  which  does 
not  exceed  the  elastic  limit  of  the  material;  then  for  tension, 

/'  =  (i  +  e)/  and  hf=(i-Xe)h  (13) 

and  similarly  for  the  case  of  compression, 

/'=(i-e)7  and  h'  =  (i  +  h)h  (13)' 

in  which  the  linear  unit-deformation  e  may  be  computed  from 
e=S/E  (Art.  9).  The  number  e  is  very  small  compared  with 
unity,  and  hence  its  square  and  higher  powers  may  be  neglected 
when  they  occur  in  algebraic  work. 


Fig.  130  Fig.  136 

Let  a  bar  of  rectangular  section  have  the  breadth  b  and  tht 
depth  d\  the  area  of  the  section  is  bd.  Under  elastic  tension, 
b  becomes  (i  —  Xe)b  and  d  becomes  (i—Ae)dt  so  that  the  area 
of  the  section  is  (i  —  2Xi)bd,  and  hence  the  decrease  in  area  is 
2\ebd,  or  2^e  is  the  decrease  per  unit  of  area.  For  example, 
let  a  bar  of  structural  steel  of  16  square  inches  cross-section  be 
stressed  in  tension  up  to  27  ooo  pounds  per  square  inch;  then 
£  =  27000/30000000=0.0009  and,  since  A  is  J,  the  decrease  in 
unit-area  is  0.0006,  so  that  the  area  of  the  section  under  this 
stress  is  16—0.0006X16  =  15.99  square  inches.  Fig.  130  illus- 
trates the  change  in  shape  for  this  case. 

The  elastic  change  in  volume  of  a  rectangular  bar  under 
tension  is  found  in  a  similar  manner.  The  length  /  increases 
to  (i  +  e)/,  the  breadth  b  decreases  to  (i—  Ae)b,  the  depth  d 
decreases  to  (i—Ae)d,  so  that  the  original  volume  Ibd  becomes 

(i  +  e)/.  (i-Ae)b.  (i-As)d=(i+e-2Jie)lbd 
This  expression  shows  that  the  volume  of  a  bar  under  tension 


34  ELASTIC  AND  ULTIMATE  DEFORMATION        CHAP.  II 

is  increased  when  A  is  less  than  J  and  this  is  the  case  for  all  ma- 
terials used  in  construction.  The  increase  in  unit-volume  is 
(i  —  2^)e,  which  becomes  %s  for  cast  iron  and  j-s  for  wrought  iron 
and  steel;  that  is,  the  increase  per  unit  of  volume  is  one  half 
or  one-third  of  the  elongation  per  unit  of  length. 

In  neglecting  the  squares  and  higher  powers  of  e  no  error 
appreciable  in  practice  has  been  committed,  as  the  student  may 
easily  see  by  numerical  instances.  Thus,  the  square  of  (i  +0.001) 
is  1.002  by  the  approximate  method,  while  its  exact  value  is 
I.C0200I.  If  v  is  a  small  number  compared  with  unity,  then 
(i  +  y)2  =  i  +  2v  and  (i  +  y)*  =  i  +  Jy;  also  i/(i  +  y)=i-y,  and 
i/(i  — y)=i  +  y.  Again,  if  y  and  /*  be  small  compared  with 
unity,  then  (i  +  y)(i  -\-JJL)  =i  +  y  +  /f  and  (i  +  y)(i  —  ju)  =i  +  y-^; 
also  (i  +  y)2(i  — 2//)2  =  i  +  2y  — 4/d.  This  principle  of  neglecting 
squares  and  cubes  is  often  of  great  value  in  approximate  numer- 
ical computations;  thus  (1.02)2  =  1.04,  and  o.9942  =  (i—  o.oo6)2  = 
i  —0.012  =0.988;  also  (0.994)*  =  (i  —  0.006) *  =  i  -0.003  =°'997' 

In  a  similar  manner  it  is  readily  shown  that  a  bar  of  length  /, 
breadth  b,  and  depth  d  under  compression  in  the  direction  of  its 
length,  has  the  area  of  its  section  increased  by  2lebd,  and  its 
volume  diminished  by  (e—2Xe)lbd,  when  e  is  the  unit-shortening 
of  length  under  a  stress  not  exceeding  the  elastic  limit  of  the 
material;  Fig.  13&  illustrates  this  case. 

The  number  A  is  called  the  '  Factor  of  Lateral  Contraction ' 
and  sometimes  also  '  Poisson's  Ratio '.  It  is  of  great  theoretic 
importance  in  many  discussions  of  the  mechanics  "of  materials, 
and  of  great  practical  value  in  the  manufacture  of  guns. 

Prob.  130.  Multiply  together  the  numbers  0.989,  1.012,  1.005  by 
the  above  approximate  method,  and  find  the  percentage  of  error  in 
the  result. 

Prob.  136.  A  bar  of  structural  steel,  2j  inches  in  diameter  and  18 
feet  6  inches  long,  is  under  a  tension  of  64  ooo  pounds.  Compute  the 
changes  in  length,  section  area,  and  volume. 

Prob.  13c.  A  bar  of  wrought  iron  is  one  square  centimeter  ID 
section  area  and  one  meter  long.  Compute  its  length  and  section  area 
when  stressed  to  its  tensile  elastic  limit. 


ART.  14 


WORK  IN  PRODUCING  DEFORMATION 


35 


ART.  14.    WORK  IN  PRODUCING  DEFORMATION 

When  a  bar  is  ta  be  put  under  tensile  stress,  the  load  is  a'pplied 
by  weights  added  in  succession,  or  by  means  of  a  pull  exerted 
by  a  machine.  In  both  cases  the  tension  increases  from  o  up 
to  P,  and  the  elongation  of  the  bar  increases  from  o  up  to  e. 
When  the  elastic  limit  is  not  exceeded,  the  elongations  are  pro- 
portional to  the  applied  forces,  and  the  relation  between  them 
may  be  represented  by  a  straight  line  as  in  Fig.  I4a.  The  work 
done  in  elongating  the  bar  is  then  the  product  of  the  mean  force 
\P  into  the  distance  e^  or  \Pe,  and  this  is  represented  in  the 
figure  by  the  area  of  the  shaded  triangle. 


Fig.  Ua 


Fig.  146 


Fig.  14c 


Let  a  be  the  section  area  of  the  bar  and  /  its  length,  S  the 
tensile  unit-stress  and  e  the  unit-elongation;  then  P=aS,  and 
e  =  £/,  and  inserting  these  in  \Pe  it  becomes  \Sz  .  al,  which  is 
the  work  done  in  elongating  the  bar  up  to  the  unit-stress  5.  Since 
al  is  the  volume  of  the  bar,  JSs  is  the  work  done  per  unit  of 
volume;  in  Fig.  146  this  work  is  represented  by  the  area  of  the 
shaded  triangle. 

The  above  investigation  applies  equally  well  to  the  case  of 
compression.  Hence,  in  any  case  of  direct  tension  or  compres- 
sion, the  work  K  of  elastic  deformation  is, 

K=\Pe  of  K=%Ss.al  (14: 

both  of  which  apply  when  the  unit-stress  S  is  not  greater  than 
the  elastic  limit  of  the  material.  This  work  is  stored  in  the  bar 
in  the  form  of  potential  stress  energy. 

As  an  example,  let  it  be  required  to  find  what  horse-powei 
engine  is  required  to  produce,  i  200  times  per  minute,  a  tension 
of  56  ooo  pounds  in  a  steel  rod  which  is  3  inches  in  diameter 
and  68  inches  long.  From  Table  16  the  section  area  is  7.07 


36 


ELASTIC  AND  ULTIMATE  DEFORMATION        CHAP.  II 


square  inches;  from  Art.  1  the  tensile  unit-stress  is  7  920  pounds 
per  square  inch;  from  Art.  10  the  unit-elongation  e  is  o.ooo  264. 
Then  from  (14)  the  work  done  in  one  application  of  the  load 
is  JX  7920X0.000264  =  1.045  inch-pounds  for  each  cubic  inch 
of  the  rod,  and  the  total  work  is  therefore  1.045X7.07X68=502 
inch-pounds.  The  horse-power  required  to  do  this  work  is 
then  502X1200/12X33  000  =  1.53. 

When  the  unit-stress  exceeds  the  elastic  limit  of  the  material, 
the  above  formulas  are  not  valid.  In  such  cases,  however,  the 
work  done  per  unit  of  volume  is  given  by  the  shaded  area  of 
the  stress  diagram  in  Fig.  14c.  This  area  may  be  approximately 
ascertained  by  dividing  it  into  trapezoids  and  computing  the 
separate  areas.  The  work  done  in  stressing  a  steel  bar  up  to 
its  ultimate  strength  is  large  compared  with  that  required  to 
stress  it  up  to  its  elastic  limit,  being  sometimes  more  than  five 
hundred  times  as  large. 

In  order  to  show  this  fact,  the  particular  case  of  a  steel  speci- 
men 4  inches  long  and  0.505  inches  in  diameter  will  be  taken. 


Load 
Pounds 

Remarks 

Stress 

Pounds  per 
Square  inch 

Elongation 
Percents. 

Partial  Work 

Inch-pounds 
per 
Cubic  Inch 

Total  Work 

Inch-pounds 
per 
Cubic  Inch 

200 

Initial  Load 

I  000 

0.00 

0 

I  000 

5  ooo 

O.OI 

'3 

o 

3.0 

3  ooo 

15  ooo 

0.04 

6.0 

3 

5  ooo 

25  ooo 

0.07 

9 

7  ooo 

35000 

0.  10 

9.0 
16.0 

18 

9  ooo 

45000 

0.14 

34 

9  600 

Elastic  Limit 

48  ooo 

o.  16 

9-3 
264.6 

44 

10  ooo 

50  ooo 

o.  70 

660  o 

308 

12  000 

60  ooo 

1.90 

968 

14  ooo 
1  6  ooo 

70  ooo 
80  ooo 

3.62 

8.50 

I   I  2O  .  O 

3  660  .  o 

2  088 

5748 

16800 

15  ooo 

Ultimate  Strength 
Rupture 

84  ooo 

75000 

15.20 

24.50 

5  494.0 
7383-0 

II  242 

18625 

The  specimen  was  placed  in  a  testing  machine  and  the  elonga- 
tions measured  corresponding  to  the  loads  in  the  first  column. 
From  these  elongations  e  the  unit-elongations  e/l  were  com- 


ART.  15  SHEARING  MODULUS  OF  ELASTICITY  37 

puted  and  the  values  of  iooe/l  are  the  percentages  of  elongation 
in  the  fourth  column,  while  the  loads  divided  by  the  section 
areas  give  the  unit-stresses  in  the  third  column.  The  elastic 
limit  was  observed  at  48000  pounds  per  square  inch  with  0.16 
percent  elongation.  The  elongation  then  rapidly  increased  with 
the  load,  as  seen  in  the  diagram  of  Fig.  4.  At  84  ooo  pounds 
per  square  inch  the  maximum  tensile  strength  was  reached  and 
the  specimen  was  elongating  rapidly.  The  load  was  then  slowly 
slackened,  but  the  elongation  continued  to  increase  very  fast 
until  rupture  occurred  at  75  ooo  pounds  per  square  inch.  The 
work  done  per  cubic  inch  of  material  may  be  approximately  com- 
puted for  any  interval  by  multiplying  the  average  unit-stress 
during  that  interval  by  the  unit-elongation  which  occurs.  Thus 
while  the  load  ranged  from  7  ooo  to  9  ooo  pounds  the  average 
unit-stress  was  40000  pounds  per  square  inch,  and  the  unit- 
elongation  was  0.0014  —  0.0010=0.0004;  hence  the  work  per 
cubic  inch  of  material  was  40  000X0.0004  =  16  inch-pounds. 
Thus  the  quantities  in  the  fifth  column  of  the  table  are  computed 
and  the  summation  of  these  gives  those  in  the  last  column. 

This  example  shows  that  the  work  done  on  the  specimen 
in  stressing  it  up  to  the  elastic  limit  was  44  inch-pounds,  while 
the  total  work  required  to  rupture  it  was  about  18  600  inch-pounds, 
both  being  per  cubic  inch  of  material.  The  volume  of  the  speci- 
men being  0.8  cubic  inches,  the  approximate  work  required  to 
rupture  it  was  14  900  inch-pounds  or  i  240  foot-pounds. 

Prob.  14a.  Compute  the  work  per  cubic  inch  which  is  done  in 
stressing  cast  iron,  wrought  iron,  and  structural  steel  up  to  their  elastic 
limits. 

Prob.  14b.  How  many  foot-pounds  of  work  are  required  to  stress 
a  wrought-iron  bar,  4  inches  in  diameter  and  54  inches  long  from 
6  ooo  pounds  per  square  inch  up  to  12  ooo  pounds  per  square  inch  ? 

ART.  15.     SHEARING  MODULUS  OF  ELASTICITY 

Let  a  body  of  constant  cross-section  and  length  /  be  subject 
to  shearing  as  shown  in  Fig.  15,  the  length  /  being  short,  so  that 
the  rectangle  A  BCD  is  deformed  into  the  rhombus  ABC'D'. 


38  ELASTIC  AND  ULTIMATE  DEFORMATION        CHAP.  II 

This  deformation  is  caused  by  the  force  P  acting  at  C  normal  to 
BC,  while  the  body  is  firmly  held  at  AB  so  that  the  points  A 

and  B  remain  immovable.  Along 
AB  there  then  acts  a  force  equal 
to  that  at  C  but  in  the  opposite 
direction,  and  the  body  is  under 
shear  from  the  action  of  these  two 
forces  (Arts.  1  and  6).  The  defor- 
mation C(7,  generally  called  the 
"detrusion",  may  be  represented 
by  e,  and  the  detrusion  per  unit  of  length  is  then  e/l.  In 
tension  and  compression  the  deformation  e  is  in  the  same  direc- 
tion as  /,  but  in  shear  it  is  normal  to  /. 

Let  a  be  the  section  area  of  the  short  bar;  then  the  shearing 
unit-stress  on  it  is  Ss=P/a.  In  tension  and  compression  this 
stress  acts  normal  to  the  section,  but  here  it  is  parallel  to  the 
section.  If  the  elastic  limit  of  the  material  is  not  exceeded  by 
the  unit-stresses,  the  unit-detrusions  are  proportional  to  them,  so 
that  the  ratio  of  P/a  to  e/l  is  a  constant,  and  this  constant  is 
called  the  'Modulus  of  Elasticity  for  Shear'.  Designating  it  by  F, 
there  results, 

P-*l±  or  *=£-.  (15) 

e/l  e 

where  e  is  the  unit-detrusion  e/l.  These  formulas  are  the  same 
as  those  of  Art.  9,  but  the  above  remarks  show  that  the  mean- 
ings of  the  letters  e  and  e  are  somewhat  different  in  the  two  cases. 

The  following  are  approximate  average  values  of  the  shear- 
ing modulus  of  elasticity  as  ascertained  by  experiments  and 
computations  which  are  explained  in  Arts.  93  and  181 : 

Material  Shearing  Modulus  of  Elasticity 

Timber,  across  grain  400  ooo  pounds  per  square  inch 

Cast  Iron  6  ooo  ooo  pounds  per  square  inch 

Wrought  Iron  10  ooo  ooo  pounds  per  square  inch 

Steel  12  ooo  ooo  pounds  per  square  inch 

For  brick,  concrete,  and  stone  there  is  but  little  known  regarding 
the  shearing  modulus  of  elasticity. 


ART.  16  HISTORICAL  NOTES  39 

By  the  help  of  the  above  formulas,  computations  regarding 
elastic  deformations  in  shear  may  be  made  in  the  same  manner 
as  those  for  tension  and  compression  in  Art.  10.  For  example, 
take  a  short  wooden  block  6  inches  long  and  3X4  inches  in  sec- 
tion subject  to  a  shear  of  24  ooo  pounds.  The  shearing  unit- 
stress  is  24000/12=2000  pounds  per  square  inch,  which  is 
probably  a  little  less  than  the  elastic  limit  for  shear  (Art.  2) 
and  hence  the  formula  applies.  The  unit-detrusion  then  is 
e  =  2  000/400  000=0.005,  and  the  total  detrusion  or  lateral  move- 
ment of  one  end  of  .the  block  with  respect  to  the  other  is 
e  =  0.005  X  6  =  0.03  inches. 

The  work  done  in  deforming  the  prism  of  Fig.  15  is  %Pe, 
when  the  load  is  applied  by  increments  so  that  the  shearing 
force  increases  from  o  up  to  P,  and  when  the  elastic  limit  is 
not  exceeded  by  the  unit-stress  S.  This  work  of  elastic  detru- 
sion may  be  written  J5e  .  a/,  which  is  identical  with  the  expres- 
sion deduced  for  tension  or  compression  in  the  last  article. 

The  above  formulas  cannot  be  used  when  the  unit-stress  S 
exceeds  the  elastic  limit  for  shear,  which  is  usually  about  75  or 
80  percent  of  that  for  tension.  They  cannot  be  used  when  the 
length  /  is  large  compared  with  the  lateral  dimensions  of  the 
body,  for  then  a  bending  occurs  so  that  EC'  is  not  a  straight 
line  and  in  this  case  e/l  cannot  be  a  constant  for  variable  values^ 
of  e  and  /.  They  are  of  principal  value  in  discussing  the  twisting 
of  shafts  (Art.  93). 

Prob.  15.  A  cast-iron  beam,  2  inches  square,  has  two  loads  of 
8  ooo  pounds  near  the  supports  as  shown  in  Fig.  6b.  The  beam 
being  originally  horizontal,  compute  the  inclination  between  a  support 
and  a  load  after  the  loads  are  placed  upon  it. 

ART.  16.    HISTORICAL  NOTES 

From  the  earliest  times  a  few  fundamental  facts  regarding 
the  strength  of  materials  were  undoubtedly  known  by  experi- 
ence, such,  for  instance,  that  stone  was  stronger  than  wood.  No 
quantitative  results  appear,  however,  to  have  been  ascertained 
and  recorded  until  about  the  middle  of  the  eighteenth  century. 


40  ELASTIC  AND  ULTIMATE  DEFORMATION        CHAP.  II 

The  first  investigation  in  the  mechanics  of  materials  seems  to 
be  that  made  by  Galileo  in  1638  on  the  flexure  of  a  beam,  in 
which  he  regarded  all  the  fibers  as  inextensible.  It  was  not 
until  more  than  a  century  afterwards  that  tests  of  the  strength 
of  bars  were  made,  and  these  related  almost  entirely  to  its  ulti- 
mate strength,  the  elastic  limit  being  unknown  and  unrecognized. 

In  1678  Hooke  announced  the  theory  of  "springy  bodies" 
which  he  expressed  by  "ut  tensio  sic  vis",  namely,  elongation 
is  proportional  to  force;  three  years  previously  he  had  performed 
an  experiment  before  the  king  of  England  which  illustrated  this 
theory,  and  in  1676  he  had  announced  it  concealed  in  the  ana- 
gram "ceiiinosssttuv."  Newton  in  1704  conjectured  that  the 
return  of  the  body  t6  its  original  form,  after  the  removal  of  the 
force,  was  caused  by  the  mutual  attraction  of  its  particles. 

Early  in  the  nineteenth  century  appeared  the  Lectures  on 
Natural  Philosophy  by  Young  in  which  the  modulus  of  elasticity 
is  for  the  first  time  introduced,  but  no  note  is  made  that  it  can  be 
deduced  or  applied  only  when  the  elastic  limit  of  the  material  is 
not  exceeded.  The  first  observations  regarding  this  limit  were 
made  a  few  years  later  by  the  experimenters  Barlow  and  Tred- 
gold,  who  noted  the  permanent  set  in  their  tests  of  cast-iron  bars. 

After  1830  metallic  bridges  came  into  use  through  the  necessi- 
ties of  railroad  construction,  and  numerous  experiments  on  cast 
iron  and  wrought  iron  were  made  in  England,  France,  and  Ger- 
many. In  1849  a  British  commission  conducted  tests  of  these 
metals  more  exhaustive  than  any  made  before,  and  for  the  first 
time  established  definite  conclusions  regarding  the  use  of  factors 
of  safety,  one  of  which  was  tha.t  the  factor  for  cast  iron  should 
be  twice  as  great  for  sudden  loads  as  for  steady  ones.  After 
this  date  specifications  for  important  structures  contained  require- 
ments regarding  working  stresses,  laboratories  for  testing  mate- 
rials  were  established,  improved  methods  and  machines  were 
devised,  and  the  theory  of  the  subject  was  greatly  extended. 

The  first  testing  machines  in  the  United  States  of  America 
were  those  built  between  1850  and  1860  for  experiments  on  gun- 
metal.  At  the  present  time  every  large  manufacturer  of  iron 


ART.  16  HISTORICAL  NOTES  41 

and  steel  has  a  testing  laboratory  where  specimens  are  con- 
stantly broken  in  order  to  gain  knowledge  whereby  the  quality 
of  the  product  may  be  improved  or  to  be  certain  that  it  meets 
the  specifications  of  the  buyers.  All  engineering  colleges  have 
testing  laboratories  for  the  instruction  of  students  and  for 
scientific  research.  The  literature  of  the  subject  is  enormous, 
and  societies,  both  national  and  international,  have  been  formed 
to  improve  the  methods  of  testing  and  to  render  more  perfect 
the  knowledge  of  the  properties  of  materials  under  stress.  All 
this  work  has  been  done  for  the  purpose  of  advancing  stability 
and  economy  in  engineering  construction,  and  the  two  principles 
stated  in  Art.  7  have  been  its  fundamental  guides. 

The  average  values  of  the  elastic  limits,  moduluses  of  elasticity, 
ultimate  strengths  and  ultimate  elongations  given  in  the  pre- 
ceding pages  and  in  Tables  2,  3,  4,  at  the  end  of  this  volume, 
have  been  derived  from  the  records  of  tests  made  since  1850. 
It  must  not  be  forgotten  that  these  values  are  subject  to  much 
variation,  depending  upon  the  quality  of  the  material,  its  method 
of  manufacture,  and  to  some  extent  upon  the  suddenness  with 
which  the  forces  are  applied.  In  practice  all  these  points  must 
be  carefully  considered  and  precise  knowledge  be  obtained,  as 
far  as  possible,  regarding  the  actual  material  in  hand.  The  next 
chapter  gives  more  detailed  information  regarding  the  elasticity 
and  strength  of  different  qualities  of  the  materials  mentioned 
in  the  preceding  pages  and  also  of  other  materials  used  in 
engineering  and  the  arts,  while  Arts.  168-170  give  discussions 
regarding  methods  of  testing. 

Prob.  16a.  Consult  Vol.  I  of  Todhunter's  History  of  the  Elasticity 
and  Strength  of  Materials  and  ascertain  the  exact  words  in  which 
Young  defined  the  Modulus  of  Elasticity. 

Prob.  166.  "Tests  of  Metals  ...  at  the  Watertown  ArsenaP  is  the 
title  of  a  publication  issued  yearly  by  the  Ordnance  Department  of 
the  U.  S.  Government.  Consult  one  of  these  volumes  and  ascertain 
the  elastic  limit  and  ultimate  strength  of  rifle-barrel  steel. 


42  MATERIALS  OF  ENGINEERING  CHAP.  Ill 

CHAPTER  III 
MATERIALS   OF  ENGINEERING 

ART.  17.     AVERAGE  WEIGHTS 

THE  following  are  average  values  of  the  weights  per  unit 
of  volume  and  of  the  specific  gravities  of  the  principal  materials 
used  in  engineering  constructions.  These  weights  should  be 
carefully  memorized  by  the  student  as  a  basis  for  more  precise 

Material  Weight                          Specific  Gravity 

Brick  125  pounds  per  cubic  foot  2.0 

Stone  1 60  pounds  per  cubic  foot  2.6 

Timber  40  pounds  per  cubic  foot  0.6 

Cast  Iron  450  pounds  per  cubic  foot  7.2 

Wrought  Iron  480  pounds  per  cubic  foot  7.7 

Steel  490  pounds  per  cubic  foot  7.8 

knowledge,  but  it  must  be  noted  that  they  are  subject  to  con- 
siderable variation  with  the  quality  of  the  material.  For  instance, 
brick  may  weigh  as  low  as  100  or  as  high  as  150  pounds  per 
cubic  foot,  according  as  it  is  soft  or  pressed.  Unless  otherwise 
stated,  the  above  average  values  will  be  used  in  all  the  follow- 
ing examples  and  problems  of  this  volume.  Table  1,  at  the  end 
of  this  volume,  gives  these  constants  in  metric  measures,  and  it 
will  be  noted  that  the  unit-weights  in  this  convenient"  system  are 
multiples  of  the  specific  gravities. 

The  following  approximate  simple  rules,  stated  in  1885  in 
the  first  edition  of  this  book,  are  in  general  use  among  engineers 
for  computing  the  weights  of  bars  and  beams  which  are  of 
uniform  cross  section: 

A  wrought-iron  bar  one  square  inch  in  section  and  one 

yard  long  weighs  ten  pounds. 

Steel  is  about  two  percent  heavier  than  wrought  iron. 
Cast  iron  is  about  six  percent  lighter  than  wrought  iron. 
Stone  is  about  one-third  the  weight  of  wrought  iron. 


ART.  is  PLASTICITY  AND  BRITTLENESS  43 

Brick  is  about  one-fourth  the  weight  of  wrought  iron. 
Timber  is  about  one-twelfth  the  weight  of  wrought  iron. 

For  example,  consider  a  bar  of  wrought  iron  1^X3  inches  and 
12  feet  long;  its  section  area  is  4.5  square  inches,  hence  its  weight 
is  45X4  =  180  pounds.  A  steel  bar  of  the  same  dimensions 
weighs  180  +  0.02X180=  about  184  pounds,  a  cast-iron  bar 
weighs  1 80 -0.06X180=  about  169  pounds,  and  a  timber  bar 
weighs  180/12  =about  15  pounds. 

By  reversing  the  above  rules,  the  section  areas  of  bars  are 
readily  computed  from  their  weights  per  yard.  Thus,  if  a  stick 
of  timber  15  feet  long  weighs  120  pounds,  its  weight  per  yard  is 
24  pounds,  and  its  section  area  is  12X2.4  =  28.8  square  inches 
approximately.  Again,  if  a  steel  bar  weighs  26.5  pounds  per 
linear  foot,  its  section  area  is  3X26.5(1-0.02)710  =  7.79  square 
inches. 

It  may  be  noted  that,  as  a  rough  general  rule,  the  strengths 
of  heavy  bodies  are  greater  than  those  of  lighter  bodies  of  the 
same  size.  Thus,  steel  is  the  heaviest  material  of  construction 
and  it  is  the  strongest.  Strictly  speaking  this  rough  rule  applies 
only  to  materials  of  a  similar  nature,  and  is  only  valid  for  com- 
paring metals  with  metals,  stony  materials  with  stony  materials, 
and  timber  with  timber.  Thus,  the  heaviest  kinds  of  timber 
are  the  strongest,  but  many  kinds  of  timber  are  stronger  than 
stone  or  brick. 

Prob.  17.  What  is  the  weight  of  a  stone  block  12X18  inches  in  sec- 
tion area  and  5  feet  long?  If  a  cast-iron  pipe  12  feet  long  weighs 
985  pounds,  what  is  its  section  area  ? 

.   ART.  18.     PLASTICITY  AND  BRITTLENESS 

The  property  of  elasticity  which  has  been  explained  in  the 
preceding  chapters  is  possessed  by  different  materials  in  differ- 
ent degrees.  A  perfectly  elastic  material  is  one  which  springs 
back  to  its  original  shape  on  the  removal  of  the  applied  force, 
this  force  having  any  magnitude  less  than  that  required  to  cause 
rupture.  No  material  with  perfect  elasticity  is  known;  india- 
rubber  is  the  nearest  approach  to  it,  but  even  this  fails  to  return 


44  MATERIALS  OF  ENGINEERING  CHAP.  Ill 

to  its  original  form  when  it  has  been  under  stress  for  some  time. 
Wrought  iron  and  steel  are  only  elastic  when  the  applied  force 
is  less  than  about  fifty  or  sixty  percent  of  that  required  to  cause 
rupture,  that  is,  when  the  unit-stress  does  not  exceed  the  elastic 
limit  (Art.  2).  Cast  iron,  stone,  and  timber  have  poorly  defined 
elastic  limits,  as  the  stress  diagrams  in  Fig.  lla  show,  and  for 
many  qualities  small  stresses  produce  permanent  sets,  so  that  their 
degree  of  elasticity  is  less  than  that  of  wrought  iron  and  steel. 

A  material  which  has  no  elasticity,  so  that  the  smallest  forces 
cause  permanent  deformations,  is  called  'plastic'.  Lead  is  an 
example  of  a  plastic  material,  for  a  flow  of  the  metal  occurs  under 
slight  forces  and  there  is  no  return  to  the  original  form.  Lead 
has  no  elastic  limit,  and  it  has  no  ultimate  compressive  strength 
in  the  ordinary  sense,  for  the  material  flows  as  the  compression 
is  applied,  so  that  the  original  prism  constantly  decreases  in  height 
and  increases  in  section  until  it  becomes  a  thin  sheet. 

Wrought  iron  and  structural  steel  are  materials  which  are 
more  or  less  plastic  under  stresses  exceeding  their  elastic  limits. 
In  tension  bars  of  these  materials  recover  a  small  part  of  their 
original  length  on  the  removal  of  the  load,  and  hence  they  have 
some  elasticity,  but  the  elongation  which  remains  as  perma- 
nent set  represents  their  plasticity.  Under  compression  the 
elasticity  is  less  and  the  plasticity  is  more  marked,  especially 
for  structural  steel,  so  that  this  variety  of  steel  cannot  be  said 
to  have  a  compressive  strength  in  the  ordinary  sense,  its  behavior 
resembling  that  of  lead. 

A  material  which  cannot  change  its  shape  without  rupture 
is  said  to  be  'brittle'.  There  is  no  body  perfectly  brittle,  but 
glass  approaches  it  most  nearly,  since  small  changes  of  shape 
are  followed  by  rupture.  Cast  iron  is  the  most  brittle  of  the 
common  materials  of  construction,  since  its  percentage  of  elon- 
gation under  tension  is  the  least  (Art.  4).  Brick  and  stone  are 
brittle  materials  compared  with  timber  or  steel.  In  general 
the  greater  the  deformation  which  a  body  will  withstand  before 
rupture  the  less  is  its  degree  of  brittleness.  Plasticity  is  the 
reverse  of  brittleness. 


ART.  18 


PLASTICITY  AND  BRITTLENESS 


45 


When  bars  are  subject  to  tensile  stress,  those  of  brittle  mate- 
rial rupture  with  but  slight  change  in  the  area  of  the  cross-sec- 
tion, while  those  of  plastic  material  rupture  with  a  considerable 
reduction  of  area  accompanied  by  a  taper  on  each  side  of  the 
place  of  rupture.  The  greater  the  reduction  of  area  (Art.  12) 
the  less  is  the  degree  of  brittleness  of  the  material.  When  prisms 
are  subject  to  compressive  stress,  those  of  brittle  material  rup- 
ture by  an  oblique  splitting  or  shearing,  as  seen  in  Fig.  185c, 
which  shows  a  cement  cube  and  an  oak  block.  The  greater  the 
brittleness  the  less  is  the  inclination  of  the  shearing  planes  to  the 
pressure  line.  Fig.  ISa  indicates  these  lines  of  rupture  for 
anthracite  coal,  Fig.  18Z>  for  paving  brick,  Fig.  I8c  for  cement 
or  concrete,  and  Fig.  I8d  for  timber;  in  all  these  cases  the  change 
in  section  area  is  very  slight.  For  a  plastic  material,  however, 
the  increase  in  section  area  is  large,  and  no  shearing  planes  arise : 
Fig.  18e  illustrates  this  case. 

I 


t        I 

Fig.  180  Fig.  186  Fig.  18c  Fig.  18d  Fig.  18e 

The  materials  of  engineering  all  possess  a  certain  degree  of 
elasticity  and  certain  degrees  of  plasticity  and  brittleness.  Hard 
steel  is  elastic  within  the  elastic  limit;  beyond  that  limit  it  is 
in  part  elastic,  in  part  plastic,  and  in  part  brittle,  while  soft  steel 
has  but  little  brittleness.  Cast  iron  has  defective  elasticity,  no 
plasticity,  and  considerable  brittleness.  Timber  resembles  cast 
iron  in  regard  to  elasticity,  but  some  varieties  are  more  plastic 
than  brittle. 

The  word  elasticity  is  used  in  the  mechanics  of  materials 
with  a  slightly  different  meaning  from  that  of  popular  language. 
In  common  parlance  a  body  is  called  elastic  when  it  can  undergo 
great  deformations  and  then  recover  its  original  form,  the  greater 
the  stretch  under  a  given  force  the  greater  being  the  elasticity. 
In  the  mechanics  of  materials  the  amount  of  the  deformation 
is  not  considered,  but  merely  the  ability  to  return  to  the  original 


46 


MATERIALS  OF  ENGINEERING 


CHAP.  Ill 


shape.  In  common  language  steel  would  not  be  spoken  of  as  an 
elastic  material,  but  according  to  the  definition  here  given  its 
degree  of  elasticity  is  a  high  one  if  the  unit-stress  does  not  exceed 
the  elastic  limit.  The  constant  called  the  modulus  of  elasticity 
(Art.  9)  is  really  an  index  of  stiffness,  that  is,  the  higher  the 
modulus  the  less  will  be  the  amount  of  deformation  under  a 
given  unit-stress. 

Prob.  180.  What  is  the  weight  in  kilograms  of  a  pig  of  lead,  0.75 
meters  long  and  215  square  centimeters  in  section  area,  its  specific 
gravity  being  11.38? 

Prob.  ISb.  A  concrete  cube  weighed  181 J  ounces  in  air  and  ioyj 
ounces  in  water.  Compute  its  specific  gravity,  and  its  weight  in  pounds 
per  cubic  foot.  What  was  the  size  of  the  concrete  cube  ? 


ART.  19.    TIMBER 

Good  timber  is  of  uniform  color  and  texture,  free  from  knots, 
sap  wood,  wind-shakes,  worm-holes,  or  decay;  it  should  also  be 
well  seasoned,  which  is  best  done  by  exposing  it  for  two  or  three 
years  to  the  weather  to  dry  "out  the  sap.  The  heaviest  timber 
is  usually  the  strongest;  also  the  darker  the  color  and  the  closer 
the  annular  rings  the  stronger  and  better  it  is,  other  things  being 
equal.  The  strength  of  timber,  except  in  the  case  of  shear,  is 
greatest  in  direction  of  the  grain,  the  sidewise  resistance  to  ten- 
sion or  compression  being  about  one  fourth  the  longitudinal. 

The   following   table   gives   average   values   of  the   ultimate 


Pounds  per  Square  Inch 

Weight 

Kind 

Pounds  per 
Cubic  Foot 

Tensile 
Strength 

Computed 
Flexure 
Strength 

Compress- 
ive 
Strength 

Hemlock 

25 

8000 

6  ooo 

5  ooo 

White  Pine 

27 

8000 

6000 

5  5°o 

Chestnut 

40 

12  OOO 

7  ooo 

5  °°° 

Red  Oak 

42 

9  ooo 

7  ooo 

6  ooo 

Yellow  Pine 

45 

15000 

II  OOO 

9  ooo 

White  Oak 

48 

12  OOO 

10  ooo 

8000 

ART.  19  TIMBER  47 

strength  of  a  few  of  the  common  kinds  of  timber  as  determined 
from  tests  of  small  specimens  which  were  carefully  selected  and 
dried.  Large  pieces  of  timber,  such  as  are  actually  used  in 
engineering  structures,  have  an  ultimate  strength  of  from  fifty 
to  eighty  percent  of  these  values.  Moreover  the  figures  are  liable 
to  a  range  of  25  percent  on  account  of  variations  in  quality  and 
condition  arising  from  place  of  growth,  time  when  cut,  and  method 
and  duration  of  seasoning.  The  amount  of  moisture  in  timber 
also  influences  its  strength;  timber  which  has  absorbed  water 
in  amount  equal  to  one-half  of  its  dry  weight  has  a  strength 
only  about  one-half  that  of  dried  timber.  To  cover  all  these 
variations  the  factor  of  safety  of  10  (Art.  7)  is  not  too  high. 

The  column  headed  'computed  flexure  strength '  in  the 
above  table  gives  the  unit-stress  computed  from  the  rupture  of 
beams,  using  the  flexure  formula  (Art.  41).  This  quantity  is 
always  intermediate  between  the  tensile  and  compressive  strengths, 
but  it  is  not  a  physical  constant,  as  will  be  explained  in  Art.  52; 
it  is  sometimes  called  the  'modulus  of  rupture '. 

The  shearing  strength  of  timber  is  still  more  variable  than 
the  tensile  or  compressive  resistance.  White  pine  across  the 
grain  may  be  put  at  2  500  pounds  per  square  inch,  and  along 
the  grain  at  500.  Chestnut  has  i  500  and  600,  yellow  pine  and 
oak  about  4  ooo  and  600  pounds  per  square  inch  in  these  direc- 
tions respectively. 

The  elastic  limit  of  timber  is  poorly  defined.  In  precise 
tests  on  good  specimens  it  is  sometimes  observed  at  about  one- 
half  the  ultimate  strength,  but  under  ordinary  conditions  it  is 
safer  to  put  it  at  one-third.  The  modulus  of  elasticity  ranges 
from  i  ooo  ooo  to  2  ooo  ooo  pounds  per  square  inch,  i  500  ooo 
being  a  good  mean  value  to  use  in  general  computations.  The 
ultimate  elongation  is  small,  usually  being  between  i  and  2  per- 
cent. Fig.  185c  shows  the  common  way  in  which  a  short  timber 
prism  ruptures  under  compression. 

The  tests  published  in  the  Census  Report  on  the  Forest  Trees 
of  North  America  (1884)  are  very  comprehensive,  as  they  include 
412  species  of  timber.  Of  these  16  species  have  a  specific  gravity 


48  MATERIALS  OF  ENGINEERING  CHAP,  in 

greater  than  i.o  and  28  species  less  than  0.4.  Even  in  the  same 
species  a  great  variation  in  weight  was  often  found;  for  instance 
white  oak  ranged  from  42  pounds  to  54  pounds  per  cubic  foot. 
The  heaviest  wood  weighed  81  pounds  per  cubic  foot  and  had 
a  compressive  strength  of  12  ooo  pounds  per  square  inch;  the 
lightest  wood  weighed  16  pounds  per  cubic  foot  and  had  a  com- 
pressive strength  of  200  pounds  per  square  inch. 

Prob.  190.  Which  is  the  safer  under  compression,  hemlock  under 
600  or  white  oak  under  800  pounds  per  square  inch  ? 

Prob.  196.  Square  sticks  of  white  pine  and  yellow  pine  are  subject 
to  a  steady  load  which  brings  a  tension  of  10  ooo  pounds  upon  each. 
Compute  their  size  for  the  proper  factor  of  safety. 

ART.  20.     BRICK 

Brick  is  made  of  clay  which  is  prepared  by  cleaning  it  care- 
fully from  pebbles  and  sand,  mixing  it  with  about  one-half  its 
volume  of  water  and  tempering  it  by  hand  stirring  or  in  a  pug- 
mill.  It  is  then  molded  into  rectangular  form  either  in  wooden 
boxes  by  hand  or  in  metal  forms  by  machines,  and  the  green 
bricks  are  placed  under  open  sheds  to  dry.  These  are  piled 
in  a  kiln  and  heated  for  nearly  two  weeks  or  until  those  nearest 
to  the  fuel  assume  a  partially  vitrified  appearance. 

Three  qualities  of  brick  are  taken  from  the  kiln :  '  arch  brick ' 
are  those  from  around  the  arches  where  the  fuel  is  burned — • 
these  are  hard  and  often  brittle;  'body  brick',  from  the  interior 
of  the  kiln,  are  of  the  best  quality;  'soft  brick',  from  the  exterior 
of  the  pile,  are  weak  and  only  suitable  for  filling.  Paving  brick 
are  burned  in  special  kilns,  often  by  natural  gas  or  by  oil,  the 
rate  of  heating  being  such  as  to  insure  toughness  and  hardness 
and  also  a  uniform  product  throughout  the  entire  kiln. 

A  common  size  is  2^X4X8^  inches,  and  the  average  weight 
4^  pounds.  A  pressed  brick,  however,  may  weigh  nearly  5^ 
pounds.  Good  bricks  should  be  of  regular  shape,  have  parallel 
and  plane  faces,  with  sharp  angles  and  edges.  They  should  be 
of  uniform  texture  and  when  struck  a  quick  blow  should  give  a 


ART.  20  BRICK  49 

sharp  metallic  ring.     The  heavier  the  brick,  other  things  being 
equal,  the  stronger  and  better  it  is. 

Poor  brick  will  absorb  when  dry  from  20  to  30  per  cent  of 
its  weight  of  water,  ordinary  qualities  absorb  from  10  to  20  per- 
cent, while  hard  paving  brick  should  not  absorb  more  than  2  or 
3  percent.  An  absorption  test  is  valuable  in  measuring  the 
capacity  of  brick  to  resist  the  disintegrating  action  of  frost,  and 
as  a  rough  general  rule  the  greater  the  amount  of  water  absorbed 
the  less  is  the  strength  and  durability. 

The  compressive  strength  of  brick  is  variable;  while  a  mean 
value  may  be  2  500  pounds  per  square  inch,  soft  brick  will 
scarcely  stand  500,  pressed  brick  may  run  to  10  ooo,  and  the 
best  qualities  of  paving  brick  have  given  15  ooo  pounds  per 
square  inch,  or  even  more.  Compressive  tests  are  generally 
made  on  half-bricks,  the  compression  being  applied  normal 
to  the  smallest  side,  that  is  parallel  to  the  width  of  the  brick; 
the  failure  usually  occurs  by  an  oblique  shearing  as  indicated 
in  Fig.  ISb. 

Brick  is  rarely  used  in  tension  and  but  little  is  known  of  its 
behavior  under  such  stress ;  the  ultimate  tensile  strength  may 
perhaps  range  from  50  to  500  pounds  per  square  inch.  The 
ultimate  shearing  strength  of  good  building  brick  is  about  500 
pounds  per  square  inch,  and  higher  for  the  paving  qualities. 

Brick  may  be  classed  as  a  brittle  material,  the  stress  dia- 
gram resembling  that  of  cast  iron.  Its  elastic  limit  is  not  well 
defined,  but  its  deformation  within  that  limit  is  greater  than 
that  of  cast  iron,  the  modulus  of  elasticity  E  generally  ranging 
from  i  ooo  ooo  to  3  ooo  ooo  pounds  per  square  inch.  Bricks 
saturated  with  water  have  about  the  same  strength  and  elastic 
properties  as  dry  bricks. 

Brick  masonry  has  a  much  smaller  strength  than  that  of 
the  individual  bricks,  on  account  of  the  lack  of  perfect  union 
at  the  joints  and  also  of  the  generally  lower  strength  of  the  mor- 
tar. The  strength  of  a  brick  pier,  indicated  by  the  loosening 
and  cracking  of  the  joints,  is  about  one-fourth  of  the  compres- 
sive strength  of  an  individual  brick. 


50  MATERIALS  OF  ENGINEERING  CHAP,  ill 

Prob.  20a.  Using  the  mean  values  of  Arts.  5  and  17  and  the  state- 
ment of  the  last  paragraph,  compute  the  height  of  a  brick  tower  which 
will  crush  at  the  base  under  its  own  weight. 


ART.  21      STONE 

Sandstone,  as  its  name  implies,  is  sand,  usually  quartzite, 
which  has  been  consolidated  under  heat  and  pressure.  It  varies 
much  in  color,  strength,  and  durability,  but  many  varieties  form 
most  valuable  building  material.  In  general  it  is  easy  to  cut 
and  dress,  but  the  variety  known  as  Potsdam  sandstone  is  very 
hard  in  some  localities. 

Limestone  is  formed  by  consolidated  marine  shells,  and  is 
of  diverse  quality.  Marble  is  limestone  which  has  been  reworked 
in  the  laboratory  of  nature  so  as  to  expel  the  impurities,  and 
leave  a  nearly  pure  carbonate  of  lime;  it  takes  a  high  polish, 
is  easily  worked,  and  makes  one  of  the  most  beautiful  building 
stones. 

Granite  is  a  rock  which  was  formerly  supposed  to  be  of  aqueous 
origin  metamorphosed  under  heat  and  pressure,  but  it  is  now 
generally  thought  to  be  of  igneous  origin ;  its  composition  is 
quartz,  feldspar,  and  mica,  but  in  the  variety  called  syenite 
the  mica  is  replaced  by  hornblende.  It  is  fairly  easy  to  work, 
is  usually  strong  and  durable,  and  some  varieties  will  take  a 
high  polish. 

Trap,  or  basalt,  is  an  igneous  rock  without  cleavage.  It  is 
hard  and  tough,  and  less  suitable  for  building  constructions 
than  other  rocks,  since  large  blocks  cannot  be  readily  obtained 
and  cut  to  size.  It  has,  however,  a  high  strength  and  is  remark- 
able for  durability. 

The  average  weights  and  ultimate  strengths  of  these  four 
classes  are  given  in  the  following  table.  These  figures  refe** 
to  small  specimens  such  as  can  be  broken  in  a  testing  machine, 
and  it  is  known  that  the  strength  of  large  blocks  per  square  inch 
is  materially  less.  The  rupture  of  a  cube  or  prism  of  stone 
under  compression  often  occurs  by  splitting,  or  rather  shearing. 


ART.  21 


STONE 


51 


in  planes  making  an  angle  of  about  25  degrees  with  the  direc- 
tion of  the  pressure  (see  Fig.  I8b). 


Kind 

Pounds  per 
Cubic  Foot 

Pounds  per  Square  Inch 

Weight 

Compressive 
Strength 

Shearing 
Strength 

Sandstone 
Limestone 
Granite 

150 
160 

7  ooo 

12  000 

I  200 

I  5OO 
2  OOO 

Trap 

175 

15  ooo 

2  500 

The  modulus  of  elasticity  of  stone  has  been  found  to  range 
from  5  ooo  ooo  to  10  ooo  ooo  pounds  per  square  inch.  The 
elastic  limit  is  difficult  to  observe,  if  indeed  any  exists,  but  the 
first  sign  of  cracking  is  sometimes  regarded  as  marking  this  limit. 
Little  is  known  regarding  the  tensile  strength,  except  that  it  is 
much  less  than  the  compressive  strength,  and  stone  is  rarely  sub- 
jected to  tensile  stresses. 

The  quality  of  a  building  stone  cannot  be  safely  inferred 
from  tests  of  strength,  as  its  durability  depends  largely  upon 
its  capacity  to  resist  the  action  of  the  weather.  Hence  corrosion 
and  freezing  tests,  impact  tests,  and  observations  of  the  behavior 
of  stone  under  conditions  of  actual  use  are  more  important  than 
the  determination  of  crushing  strength  in  a  compression  machine. 
The  strength  of  a  stone  pier  is  only  about  one-fourth  of  that 
of  the  stone  itself,  on  account  of  the  weakness  of  the  mortar 
joints. 

Slate  is  an  argillaceous  stone  consolidated  under  very  heavy 
pressure  which  has  produced  a  marked  cleavage.  It  is  split 
into  plates  J  inch  in  thickness  for  use  as  roofing  slate,  and  in 
larger  blocks  is  used  for  pavements  and  steps.  Its  weight  per 
cubic  foot  is  about  175  pounds,  its  compressive  strength  about 
10  ooo  pounds  per  square  inch,  and  its  computed  flexural  strength 
about  7  ooo  pounds  per  square  inch.  Slate  is  liable  to  corrode 
under  the  action  of  the  atmosphere,  and  its  marked  cleavage 
and  grain  render  its  strength  variable  in  different  directions; 
hence  it  is  unsuitable  for  a  building  stone. 


52 


MATERIALS  OF  ENGINEERING 


CHAP.  Ill 


Prob.  21.  How  many  cubic  yards  of  masonry  are  contained  in  a 
pier  12X30  feet  at  the  base,  8X24  feet  at  the  top,  and  i6j  feet  high? 
What  is  its  cost  at  $6.37  per  cubic  yard? 

ART.  22.    MORTAR  AND  CONCRETE 

Common  mortar  is  composed  of  one  part  of  lime  to  five  parts 
of  sand  by  measure.  When  six  months  old  its  tensile  strength 
is  from  15  to  30  and  its  compressive  strength  from  150  co  300 
pounds  per  square  inch.  Its  strength  slowly  increases  with  age, 
and  a  smaller  proportion  of  sand  produces  mortar  stronger  than 
the  above  values. 

Hydraulic  mortar  is  composed  of  hydraulic  cement  and 
sand  in  varying  proportions.  The  less  the  proportion  of  sand 
the  greater  is  its  strength.  If  S  be  the  strength  of  neat  cement, 
that  is,  of  cement  with  no  sand,  the  strength  of  mortar  having 
p  parts  of  sand  to  one  part  of  cement  is  about  S/(i+p)  as  a 
rough  approximation.  A  common  proportion  is  3  parts  of  sand 
to  i  of  cement,  the  strength  of  this  being  about  one-fourth  that 
of  the  cement  itself.  The  strength  of  hydraulic  mortar  also 
increases  with  its  age. 

There  are  two  classes  of  hydraulic  cements,  the  natural 
cements  and  the  Portland  cements.  The  former  are  of  lighter 
color,  lower  weight,  and  lesser  strength  than  the  latter,  but  they 
are  quicker  in  setting  and  cheaper  in  price.  The  following 
table  gives  average  ultimate  tensile  strengths  in  pounds  per 


Proportion 
of  Sand 

NATURAL 

PORTLAND 

One 

Week 

One 

Month 

One 
Year 

One 
Week 

One 

Month 

One 
Year 

0 

ISO 

250 

400 

500 

680 

700 

I  to  I 

no 

20O 

300 

330 

400 

500 

2  to  I 

90 

15° 

240 

240 

310 

410 

3  to  i 

5° 

IOO 

190 

1  80 

250 

35° 

4  to  i 

3o 

75 

IOO 

IOO 

1  80 

260 

square  inch  of  mortars  of  both  classes  of  different  ages  and  differ- 
ent proportions  of  sand.     These  figures  are  obtained  from  tests 


ART.  22  MORTAR   AND    CONCRETE  53 

of  briquettes  carefully  made  in  molds,  and  are  higher  than 
hydraulic  mortar  made  under  usual  circumstances  will  give. 
The  briquette  is  removed  from  the  mold  when  set,  allowed  to 
remain  in  air  for  one  day,  and  then  put  under  water  for  the 
remainder  of  the  time. 

Since  cements  and  mortars  are  used  only  in  compression,  the 
tensile  tests  may  seem  an  inappropriate  one.  Compressive  tests, 
however,  are  more  expensive  than  tensile  ones.  It  may  be  taken 
as  a  general  rule  that  the  compressive  strength  of  a  material 
increases  with  the  tensile  strength,  and  it  is  certain  that  the 
universal  adoption  of  the  tensile  tests  has  done  much  to  greatly 
improve  the  quality  of  hydraulic  cements. 

For  neat  cement  a  one-day  tensile  test  is  frequently  employed, 
the  briquette  being  put  under  water  as  soon  as  it  has  set.  For 
natural  cement  a  briquette  one  square  inch  in  section  should 
give  a  tensile  strength  of  75  pounds,  while  one  of  Portland  cement 
should  give  125  pounds.  To  secure  these  results,  however, 
the  material  should  be  thoroughly  rammed  into  the  molds, 
no  more  water  being  used  than  necessary,  and  the  quality  of 
the  cement  must  be  good. 

The  compressive  strength  of  hydraulic  cements  and  mortars 
is  from  six  to  ten  times  the  tensile  strength.  Fig.  169&  shows 
the  manner  in  which  a  cube  of  cement  fails  under  compression. 
Neat  Portland  cement  when  one  month  old  has  a  compressive 
strength  of  about  3  ooo,  and  when  one  year  old  about  5  ooo 
pounds  per  square  inch.  Natural  cement  mortar  when  three 
or  four  years  old  has  a  compressive  strength  of  about  2  500  pounds 
per  square  inch.  The  adhesion  of  cement  to  stone  or  brick  is 
somewhat  less  than  the  tensile  strength.  The  shearing  strength 
of  cement  or  mortar  is  much  less  than  the  compressive  strength, 
usually  only  about  one-fourth  or  one-third. 

The  strength  of  cement  and  mortar  is  influenced  by  many 
causes:  the  quality  of  the  stone  or  materials  from  which  the 
cement  is  made,  the  method  of  manufacture,  the  age  of  the 
cement,  the  kind  of  sand,  the  method  of  mixing,  and  even  the 
amount  of  water  used.  Mortar  joints  are,  as  a  rule,  weaker 


54  MATERIALS  OF  ENGINEERING  CHAP,  ill 

than  the  bricks  or  stones  which  they  unite,  and  the  failure  of  a 
masonry  wall  usually  begins  by  cracking  along  these  joints. 

Concrete  is  an  artificial  stone  which  is  made  by  mixing 
hydraulic  mortar  and  broken  stones.  The  best  proportions 
are  such  that  the  grains  of  sand  fill  all  the  voids  between  the 
stones,  while  the  cement  fills  all  the  voids  between  the  sand  grains, 
as  also  those  between  the  sand  and  the  stones.  Common  pro- 
portions for  a  first-class  concrete  are  i  cement,  2  sand,  and  4 
broken  stone,  by  measure.  Concrete  is  mainly  used  for  floors, 
walls,  foundations,  and  monolithic  structures,  but  sometimes 
blocks  are  made  which  are  laid  together  like  stone  masonry. 
Its  use  has  greatly  increased  since  1900  and  many  concrete  piers 
and  bridges  have  been  built.  Reinforced  concrete  is  concrete 
in  which  steel  rods  are  imbedded  in  order  to  increase  its  tensile 
resistance  (Art.  113). 

The  average  weight  of  concrete  is  about  150  pounds  per 
cubic  foot.  Its  strength  increases  with  age,  like  that  of  hydraulic 
mortar.  For  concrete  made  with  the  proportions  i  cement, 

2  sand,  4  broken  stone,  when  six  months  old,  the  average  ten- 
sile strength  is  400  pounds  per  square  inch,  compressive  strength 

3  500  pounds  per  square  inch,  and  modulus  of  elasticity  3  ooo  ooo 
pounds   per   square   inch.     For   concrete   with   the   proportions 

1  cement,  3  sand,  6  broken  stone,  when  six  months  old,  the  aver- 
age tensile  strength  is  300  pounds  per  square  inch,  compressive 
strength  2  500  pounds  per  square  inch,  and  modulus  of  elasticity 

2  ooo  ooo  pounds  per  square  inch.     These  figures  refer  to  con- 
crete in  which  Portland  cement  is  used,  the  strength  being  about 
twenty  percent  less  for  natural  cement  concrete.     When  one  year 
old  the  ultimate  strengths   are  a  little   greater  than   the  abov? 
figures.     The  elastic  limit  of  concrete  is  not  well  defined,  and 
if  one  is  to  be  stated,  that  in  compression  may  be  roughly  put 
at  about  one-third  or  one-fourth  of  the  compressive  strength. 
The  allowable  unit-stress  usually  ranges  from  one-sixth  to  one- 
eighth  of  the  ultimate  strength. 

Prob.  22.  Let  the  pier  of  Problem  21  be  concrete  and  carry  a  load 
of  460  ooo  pounds.  Compute  its  weight  and  the  compressive  unit- 
stress  on  its  base. 


A.RT.  23  CAST  IRON  55 


ART.  23.     CAST  IRON 

Cast  iron  is  a  modern  product,  having  been  first  made  in 
England  about  the  beginning  of  the  fifteenth  century.  Ores 
of  iron  are  melted  in  a  blast  furnace,  producing  pig  iron.  The 
pig  iron  is  remelted  in  a  cupola  furnace  and  poured  into  molds, 
thus  forming  castings.  Beams,  columns,  pipes,  braces,  and 
blocks  of  every  shape  required  in  engineering  structures  are  thus 
produced. 

Pig  iron  is  divided  into  two  classes,  Foundry  pig  and  Forge 
pig,  the  former  being  used  for  castings  and  the  latter  for  making 
wrought  iron  and  steel.  Foundry  pig  has  a  dark-gray  fracture, 
with  large  crystals  and  a  metallic  luster;  forge  pig  has  a  light- 
gray  or  silver-white  fracture,  with  small  crystals.  Foundry 
pig  has  a  specific  gravity  of  from  7.1  to  7.2,  and  it  contains  from 
6  to  4  percent  of  carbon ;  forge  pig  has  a  specific  gravity  of  from 
7.1  to  7.4,  and  it  contains  from  4  to  2  percent  of  carbon.  The 
higher  the  percentage  of  carbon  the  less  is  the  specific  gravity, 
and  the  easier  it  is  to  melt  the  pig.  Besides  the  carbon  there 
are  present  from  i  to  5  percent  of  other  impurities,  such  as 
silicon,  manganese,  and  phosphorus. 

The  properties  and  strength  of  castings  depend  upon  the 
quality  of  the  ores  and  the  method  of  their  manufacture  in  both 
the  blast  and  the  cupola  furnace.  Cold-blast  pig  produces 
stronger  iron  than  the  hot-blast,  but  it  is  more  expensive.  Long- 
continued  fusion  improves  the  quality  of  the  product,  as  also 
do  repeated  meltings.  The  darkest  grades  of  foundry  pig  make 
the  smoothest  castings,  but  they  are  apt  to  be  brittle;  the  light- 
gray  grades  make  tough  castings,  but  they  are  apt  to  contain 
blow-holes  or  imperfections. 

The  percentage  of  carbon  in  cast  iron  is  a  controlling  fact  or 
which  governs  its  strength,  particularly  that  percentage  which 
is  chemically  combined  with  the  iron.  For  example,  the  fol- 
lowing are  the  results  of  tests  made  by  Wade  about  1860  of  three 
classes  of  cast  iron  then  used  for  guns,  the  tensile  strength  being 
in  pounds  per  square  inch : 


56  MATERIALS  OF  ENGINEERING  CHAP,  in 

XT  c       "c    r>       -I          Percentage  of  Carbon  Ultimate 

No.        Specific  Gravity     Gnp}dte          Combined     Tensile  Strength 

1  7.204  2.06  1.78  28  800 

2  7-T54  2-3°  J-46  2480x3 

3  7-°%7  2-83  0.82  20  TOO 

Here  it  is  seen  that  the  total  carbon  is  about  the  same  in  the 
three  kinds,  but  the  greater  the  percentage  of  combined  carbon 
the  higher  is  the  specific  gravity  and  ultimate  strength. 

As  average  values  for  the  ultimate  strength  of  cast  iron, 
20  ooo  and  90  ooo  pounds  per  square  inch  in  tension  and  com- 
pression respectively  are  good  figures.  In  any  particular  case, 
however,  a  variation  of  from  10  to  20  per  cent  from  these  values 
may  be  expected,  owing  to  the  variations  in  quality.  For  first- 
class  gun  iron  Wade  found  a  tensile  strength  of  over  30  ooo 
and  a  compressive  strength  of  over  1 50  ooo  pounds  per  square 
inch.  On  the  other  hand  medium-quality  castings  often  have 
a  tensile  strength  less  than  16  ooo  pounds  per  square  inch. 

In  tensile  tests  the  elongations  within  the  elastic  limit  are 
not  exactly  proportional  to  the  applied  loads,  so  that  the  elastic 
limit  is  poorly  defined,  and  the  modulus  of  elasticity  may  range 
from  12000000  to  18000000  pounds  per  square  inch.  The 
elastic  limit  in  tension  may  be  roughly  stated  at  6  ooo  pounds 
per  square  inch,  but  that  for  compression  is  much  higher,  prob- 
ably about  20  ooo  pounds  per  square  inch. 

The  flexural  test  is  a  good  one  for  comparing  the  strength 
of  different  bars  of  cast  iron.  A  bar  one  inch  square  and  14 
inches  long  is  laid  on  two  supports  12  inches  apart  and  a  load 
gradually  applied  to  the  middle  until  rupture  occurs;  the  load 
required  will  be  about  2  ooo  pounds  for  average  grades.  For 
a  bar  2X1  inches  in  section  area,  laid  flatwise  on  two  supports 
24  inches  apart,  the  load  at  the  middle  which  causes  rupture 
will  also  be  about  2  ooo  pounds. 

The  high  compressive  strength  and  the  cheapness  of  cast 
iron  render  it  a  valuable  material  for  many  purposes,  but  its 
brittleness,  low  tensile  strength,  and  lack  of  ductility  forbid 
its  use  5n  structures  subject  to  variations  of  load  or  to  shocks, 


ART.  24  WROUGHT  IRON  57 

Its  ultimate  elongation  being  scarcely  one  percent,  the  work 
required  to  cause  rupture  is  small  compared  to  that  for  wrought 
iron  and  steel.  Cast  iron  is  no  longer  used  in  bridges,  and  for 
important  buildings  it  has  been  but  little  employed  since  1895. 

Malleable  cast  iron  is  made  by  surrounding  castings  with 
decarbonizing  material  and  heating  them  in  annealing  ovens. 
The  carbon  is  thus  partly  removed  to  a  certain  depth  below 
the  surface  of  the  castings,  and  this  renders  the  material  much 
stronger  in  tension,  while  its  elongation  is  increased.  The  process 
is  only  applicable  to  small  castings,  but  for  these  the  tensile  strength 
may  be  doubled  and  the  ultimate  elongation  be  made  eight  or 
ten  times  as  great  as  that  of  common  cast  iron,  thus  rendering 
it  more  efficient  in  resisting  shocks. 

Prob.  23.  Compute  the  weight  of  a  cast-iron  water  pipe  12  feet  long 
and  20  inches  in  internal  diameter,  the  average  thickness  of  the  metal 
being  ij  inches. 

ART.  24.    WROUGHT  IRON 

The  ancient  peoples  of  Europe  and  Asia  were  acquainted 
with  wrought  iron  and  steel  to  a  limited  extent.  The  name  of 
an  ironmaster  is  mentioned  in  Genesis  iv  22,  and  in  one  of 
the  oldest  pyramids  of  Egypt  a  piece  of  iron  has  been  found. 
It  was  produced,  undoubtedly,  by  the  action  of  a  hot  fire  on 
very  pure  ore.  The  ancient  Britons  built  bloomaries  on  the 
tops  of  high  hills,  a  tunnel  opening  toward  the  north  furnishing 
a  draft  for  the  fire  which  caused  the  carbon  and  other  impurities 
to  be  expelled  from  the  ore,  leaving  behind  nearly  pure  metallic 
iron. 

Modern  methods  of  manufacturing  wrought  iron  are  mainlv 
by  the  use  of  forge  pig  (Art.  23)  in  the  puddling  process.  Here 
the  forge  pig  is  subjected  to  the  oxidizing  flame  of  a  blast  in  a 
reverberatory  furnace,  where  it  is  formed  into  pasty  balls  by  the 
puddler.  A  ball  taken  from  the  furnace  is  run  through  a  squeezer 
to  expel  the  cinder  and  then  rolled  into  a  muck  bar.  The  muck 
bars  are  cut,  laid  in  piles,  heated,  and  rolled,  forming  what  is 
called  merchant  bar.  If  this  is  cut,  piled,  and  rolled  again  a 


58  MATERIALS  OF  ENGINEERING  CHAP,  ill 

better   product    called   best   iron   is   produced.     A  third  rolling 
gives  ' best-best'  iron,  a  superior  quality,  but  high  in  price. 

The  product  of  the  rolling  mill  is  bar  iron,  plate  iron,  shape 
iron,  beams,  and  rails.  Bar  iron  is  round,  square,  and  rectan- 
gular in  section ;  plate  iron  is  from  J  to  i  inch  thick,  and  of  vary- 
ing widths  and  lengths ;  shape  iron  includes  angles,  tees,  channels, 
and  other  forms  used  in  structural  work;  beams  are  I-shaped, 
and  of  the  deck  or  bulb  form  (Art.  44). 

Wrought  iron  is  metallic  iron  containing  less  than  0.25  per- 
cent of  carbon,  and  which  has  been  manufactured  without  cast- 
ing from  the  fluid  state.  Its  elastic  limit  is  well  defined  at  about 
24  ooo  pounds  per  square  inch  and  the  yield  point  at  about 
26  ooo  pounds  per  square  inch;  within  this  limit  it  is  stiffer  than 
cast  iron,  the  modulus  of  elasticity  being  about  25  ooo  ooo  pounds 
per  square  inch.  Its  average  ultimate  tensile  strength  is  about 
50  ooo  pounds  per  square  inch,  and  its  ultimate  elongation  is 
from  25  to  30  percent.  In  comparison  it  is  plastic  beyond  the 
elastic  limit,  so  that  the  compressive  strength  given  in  Art.  5 
has  but  little  meaning.  It  is  malleable,  can  be  forged  and  welded, 
and  has  a  high  capacity  to  withstand  the  action  of  shocks;  it 
cannot,  however,  be  tempered  so  as  to  form  cutting  tools. 

The  cold-bend  test  for  wrought  iron  is  an  important  one  for 
judging  of  general  quality.  A  bar,  perhaps  fxf  inches  and 
15  inches  long,  is  bent  when  cold  either  by  pressure  or  by  blows 
of  a  hammer.  Bridge  iron  should  bend,  without  cracking,  through 
an  angle  of  90  degrees  to  a  curve  whose  radius  is  twice  the  thick- 
ness of  the  bar.  Rivet  iron  should  bend,  without  showing  signs 
of  fracture,  through  180  degrees  until  the  sides  of  the  bar  are  in 
contact.  Wrought  iron  that  breaks  under  this  test  is  lacking 
in  both  strength  and  ductility. 

The  process  of  manufacture,  as  well  as  the  quality  of  the 
pig  iron,  influences  the  strength  of  wrought  iron.  The  higher 
the  percentage  of  carbon  the  greater  is  the  strength.  Best  iron 
is  10  percent  stronger  than  ordinary  merchant  iron  owing  to  the 
influence  of  the  second  rolling.  Cold  rolling  causes  a  marked 
increase  in  elastic  limit  and  ultimate  strength,  but  a  decrease 


ART.  24  WROUGHT  IRON  59 

in  ductility  or  ultimate  elongation.  Annealing  lowers  the  ulti- 
mate strength,  but  increases  the  elongation.  Iron  wire,  owing 
to  the  process  of  drawing,  has  a  high  tensile  strength,  sometimes 
greater  than  100  ooo  pounds  per  square  inch. 

Good  wrought  iron  when  broken  by  tension  shows  a  fibrous 
structure.  If,  however,  it  be  subject  to  shocks  or  to  repeated 
stresses  which  exceed  the  elastic  limit,  the  molecular  structure 
becomes  changed  so  that  the  fracture  is  more  or  less  crystalline. 
The  effect  of  a  stress  exceeding  the  elastic  limit  is  to  cause  a 
permanent  set,  but  the  elastic  limit  will  be  found  to  be  higher 
than  before.  This  is  decidedly  injurious  to  the  material  on 
account  of  the  accompanying  change  in  structure  and  because 
its  capacity  to  resist  work  is  less  than  before,  as  Fig.  116  shows; 
hence  it  is  a  fundamental  principle  that  the  working  unit-stresses 
should  not  exceed  the  elastic  limit.  For  proper  security  the 
allowable  unit-stress  should  seldom  be  greater  than  one-half 
the  elastic  limit. 

Prior  to  1890  wrought  iron  was  generally  employed  for  struc- 
tural purposes,  but  medium  steel,  on  account  of  its  smaller  cost 
and  greater  strength,  then  began  to  come  into  use,  and  the  change 
was  so  rapid  that  since  1900  it  has  almost  entirely  displaced 
wrought  iron  in  bridge  and  building  construction.  Wrought  iron 
is,  however,  still  employed  for  chains,  for  stay  bolts,  and  for  many 
other  purposes  where  great  ductility  and  toughness  are  demanded. 
Owing  to  the  fact  that  it  was  used  so  extensively  in  the  nineteenth 
century  and  that  its  properties  .were  then  so  well  ascertained  by 
numerous  tests,  it  will  long  remain  in  engineering  literature  as 
a  standard  with  which  other  materials  may  be  advantageously 
compared.  Since  1900  much  pig  iron  has  been  made  in  America 
by  casting  in  open  molds  instead  of  in  a  sand  bed,  and  wrought 
^ron  made  from  such  pig  appears  to  be  lower  in  strength  than  that 
of  the  nineteenth  century. 

Prob.  24.  Compute  the  section  area  of  a  wrought-iron  bar  30  feet 
long  which  weighs  418  pounds.  If  this  bar  is  hung  vertically  at  its 
upper  end,  compute  the  unit-stress  at  that  end  and  at  the  middle. 
What  load,  hung  at  the  lower  end,  will  stress  it  to  the  elastic  limit? 


60  MATERIALS  OF  ENGINEERING  CHAP,  m 

ART.  25.     STEEL 

Steel  was  originally  produced  directly  from  pure  iron  ore  by 
the  action  of  a  hot  fire,  which  did  not  remove  the  carbon  to  a 
sufficient  extent  to  form  wrought  iron.  The  modern  processes, 
however,  involve  the  fusion  of  the  ore,  and  the  definition  of 
the  United  States  law  is  that  "  steel  is  iron  produced  by  fusion 
by  any  process,  and  which  is  malleable."  Chemically,  steel  is  a 
compound  of  iron  and  carbon  generally  intermediate  in  com- 
position between  cast  and  wrought  iron,  but  having  a  higher 
specific  gravity  than  either.  The  following  comparison  points 
out  the  distinctive  differences  between  the  three  kinds  of  iron: 

Percent  of      Specific  Pronerties 

Carbon         Gravity 

Cast  iron,  5       to  2          7.2     Not  malleable,  not  temperable 

Steel,  i.5otoo.io     7.8     Malleable  and  temperable 

Wrought  iron,    0.30  to  0.05     7.7     Malleable,  not  temperable 

It  should  be  observed  that  the  percentage  of  carbon  alone  is 
not  sufficient  to  distinguish  steel  from  wrought  iron;  also,  that 
the  mean  values  of  specific  gravity  stated  are  in  each  case  sub- 
ject to  considerable  variation;  further,  only  the  hard  steels  are 
temperable,  the  softer  grades  resembling  wrought  iron. 

The  three  principal  methods  of  manufacture  are  the  cruci- 
ble process,  the  open-hearth  process,  and  the  Bessemer  process. 
In  the  crucible  process  impure  wrought  iron  or  blister  steel, 
with  carbon  and  a  flux,  are  fused  in  a  sealed  vessel  to  which 
air  cannot  obtain  access;  the  best  tool  steels  are  thus  made. 
In  the  open-hearth  process  pig  iron  is  melted,  wrought-iron 
scrap  being  added  until  the  proper  degree  of  carbonization  is 
secured.  In  the  Bessemer  process  pig  iron  is  completely  decar- 
bonized in  a  converter  by  an  air  blast  and  then  recarbonized 
to  the  proper  degree  by  the  addition  of  spiegeleisen.  The  metal 
from  the  open-hearth  furnace  or  from  the  Bessemer  converter 
is  cast  into  ingots  which  are  rolled  in  mills  to  the  required  forms. 
The  open-hearth  process  produces  steel  for  machines,  shafts, 
axles,  springs,  armor  plates,  and  for  structural  purposes;  the 
Bessemer  process  mainly  produces  steel  for  railroad  rails. 


ART.  25  STEEL  61 

'Acid  steel '  is  that  made  in  a  furnace  having  a  silicious  lining, 
while' 'basic  steel'  is  that  made  in  a  furnace  with  dolomitic  linings. 
These  terms  have  no  reference  to  quality  and  refer  mainly  to 
process  of  manufacture,  but  the  basic  process  enables  ores  high 
in  phosphorus  to  be  used,  the  phosphorus  being  removed  by 
the  addition  of  lime.  Acid  steel  is  slightly  stronger  than  basic 
steel,  owing  partly  to  the  higher  percentage  of  phosphorus  and 
partly  to  the  effect  of  the  lime  in  forming  slag  in  the  basic  steel. 
Since  1900  more  than  three-fourths  of  the  open-hearth  steel  pro- 
duced in  the  United  States  has  been  basic;  the  Bessemer  product 
on  the  other  hand  being  entirely  acid. 

The  physical  properties  of  steel  depend  both  upon  method 
of  manufacture  and  chemical  composition,  the  carbon  having 
the  controlling  influence  upon  strength.  Phosphorus  increases 
strength,  but  it  promotes  brittleness;  manganese  increases 
strength  in  a  less  degree,  and  it  promotes  malleability;  sulphur 
causes  red-shortness,  or  a  tendency  of  the  steel  to  crumble  while 
being  rolled;  and  silicon  increases  hardness. 

Many  formulas  have  been  deduced   to   exhibit   the  relation 
between  the  tensile  strength  and  the  chemical  composition,  but 
none  of  these  applies  to  all  classes  of  steel.     The  rough  rules, 
£,=45  000+ 108  oooC  St=4S  000+90  oooC 

give  an  approximate  idea  of  the  influence  of  carbon  in  acid  and 
basic  unannealed  open-hearth  steel  respectively,  C  being  the 
percentage  of  carbon  and  St  the  tensile  strength  in  pounds  per 
square  inch.  Thus,  acid  steel  with  0.40  percent  of  carbon  has  a 
tensile  strength  of  about  88  ooo  pounds  per  square  inch,  while 
basic  steel  has  about  81  ooo  pounds  per  square  inch.  When  the 
percentages  of  phosphorus  and  manganese  are  also  known,  the 
following  formulas  which  have  been  deduced  from  the  exhaustive 
discussion  given  by  Campbell  in  1905  may  be  used  to  give 
more  reliable  results,  namely, 

St=  40  000+  68  oooC+ 100  oooP+  80  oooCM 

£,=  38  800+65  oooC+ 100  oooP+9  oooM+4o  oooCM 

the  first  being  for  acid  and  the  second  for  basic  open-hearth 


62  MATERIALS  OF  ENGINEERING  CHAP,  in 

steel.  Here  C  is  the  percentage  of  carbon,  P  that  of  phosphorus, 
M  that  of  manganese,  and  St  the  tensile  strength  in  pounds  per 
square  inch.  For  example,  acid  steel  having  0.344  percent  of 
carbon,  0.045  percent  of  phosphorus,  and  0.70  per  cent  of 
manganese  has  a  tensile  strength  of  87  200  pounds  per  square 
inch;  basic  steel  having  0.344  percent  of  carbon,  0.020  percent 
of  phosphorus,  and  0.35  percent  of  manganese  has  a  tensile 
strength  of  71  100  pounds  per  square  inch.  These  formulas  do 
not  apply  to  steel  with  a  percentage  of  carbon  higher  than  0.75. 

Carbon  is  the  controlling  element  in  regard  to  strength,  and 
the  same  is  the  case  with  respect  to  ultimate  elongation.  The 
higher  the  percentage  of  carbon,  within  a  reasonable  limit,  the 
greater  is  the  strength  and  the  less  the  ultimate  elongation.  The 
product  of  strength  and  elongation  is  approximately  constant, 
and  hence  the  ultimate  elongation  is  approximately  inversely 
proportional  to  the  tensile  strength.  A  rule  frequently  given  is 
that  the  percentage  of  elongation  equals  i  500  ooo/St ;  thus, 
for  a  tensile  strength  of  80  ooo  pounds  per  square  inch  the  ulti- 
mate elongation  is  about  19  percent.  This  rule,  however,  gives 
too  high  elongations  for  very  strong  steel. 

A  classification  of  steel  according  to  the  percentage  of  car- 
bon which  it  contains  and  its  capacity  for  taking  temper  or  being 
welded,  is  as  follows : 

Soft,  0.05-0. 2oC,  not  temperable,  easily  welded 

Medium,  0.15-0.40(7,  poor  temper,  weldable 

Hard,  0.30-0. 7oC,  temperable,  welded  with  difficulty 

Very  hard,  o.6o-i.ooC,  high  temper,  not  weldable 

It  is  seen  that  these  classes  overlap  so  that  there  is  no  distinct 
line  of  demarcation,  and  in  fact  the  words  soft,  medium,  and 
hard,  are  frequently  used  without  precision  and  only  for  com- 
parative purposes.  The  term  'strong  steel'  in  the  two  pre- 
ceding chapters  has  been  introduced  for  educational  purposes 
only,  in  order  to  divide  steel  into  two  classes  for  the  benefit  of 
beginners  in  engineering. 

Steel  is  frequently  classified,  with  reference  to  its  uses,  and 
the  following  is  such  a  classification  giving  average  elastic  limits 


ART.  25  STEEL  63 

and  ultimate  tensile  strengths  in  pounds  per  square  inch  and 
ultimate  elongations  in  percentages.  For  the  elastic  limit  a 
variation  of  about  2  ooo  and  for  the  ultimate  strength  a  varia- 

Elastic  Limit     Tensile  Strength     Elongation 
Structural  steel 

for  rivets  30  ooo  55  ooo  30 

for  beams  and  shapes  35  ooo  60  ooo  27 
Boiler  steel 

for  rivets  25  ooo  50  ooo  30 

for  plates  30  ooo  60  ooo  26 

Machinery  steel  40  ooo  75  ooo  20 

Gun  steel  50  ooo  90  ooo  18 

Axle  steel  55000  100  ooo  15 

Spring  steel  60000  125000  12 

Cable  wire  steel  100  ooo  200  ooo  8 

tion  of  4  coo  or  5  ooo  pounds  per  square  inch  from  these  mean 
values  may  be  expected.  The  ultimate  elongations  are  subject 
to  marked  variation  according  to  the  ratio  of  the  length  of  the 
test  specimen  to  its  diameter;  those  here  given  are  for  the  standard 
8-inch  specimen  (Art.  169)..  In  Fig.  169#  are  shown  an  unbroken 
and  two  broken  specimens  of  the  2 -inch  size  which  has  been 
much  used  since  1900;  this  gives  higher  percentages  of  elongation 
than  the  8-inch  specimen. 

The  soft  and  medium  steels  resemble  wrought  iron  in  having 
a  yield  point  (Art.  11)  which  is  from  2  ooo  to  4000  pounds  per 
square  inch  above  the  elastic  limit,  while  very  hard  steels  have 
no  yield  point,  as  the  stress  diagrams  in  Fig.  lla  show.  The 
elastic  limit  in  tension  is  a  little  higher  than  one-half  of  the  uti- 
mate  strength.  The  modulus  of  elasticity  is  subject  to  but  little 
variation  with  the  percentage  of  carbon,  and  the  mean  value  of 
30  ooo  ooo  pounds  per  square  inch  may  be  used  in  computations 
for  both  tensile  and  compressive  stresses  that  do  not  exceed  the 
elastic  limit.  The  modulus  of  elasticity  for  shearing  is  about 
three-eighths  of  that  for  tension  and  compression.  Soft  and 
medium  steel  will  withstand  a  cold-bend  test  similar  to  that 
described  in  the  last  article,  but  some  of  the  hard  steels  will  fail 
to  do  §o  on  account  of  their  lack  of  ductility. 


64  MATERIALS  OF  ENGINEERING  CHAP,  ill 

The  compressive  strength  of  the  soft  steels  may  be  said  to 
be  about  the  same  as  the  tensile  strength,  since  when  this  pres- 
sure is  reached  the  shortened  specimen  is  badly  cracked.  The 
soft  steels  resemble  wrought  iron  in  being  plastic  under  com- 
pressive stress  exceeding  the  elastic  limit,  and  some  authorities 
regard  the  yield  point  as  the  compressive  strength.  Hard  steels, 
on  the  other  hand,  are  not  plastic  beyond  the  elastic  limit,  but 
their  behavior  is  like  that  of  brittle  materials  (Art.  18).  The 
compressive  strength  of  the  hard  steels  is  much  higher  than  the 
tensile  strength  and  in  this  respect  they  resemble  cast  iron;  the 
greatest  value  recorded  is  392  ooo  pounds  per  square  inch.  The 
shearing  strength  of  steel  is  usually  about  eighty  percent  of  the 
tensile  strength. 

The  strength  of  steel  may  be  greatly  increased  by  the  processes 
of  forging  and  drawing.  Forging  under  a  hammer  or  press 
renders  the  material  more  compact  and  increases  both  specific 
gravity  and  strength.  The  process  of  drawing  steel  bars  into 
wire  has  a  similar  result,  and  wire  has  been  made  having  a  ten- 
sile strength  of  250  ooo  pounds  per  square  inch,  while  the  wire 
used  for  the  cables  of  suspension  bridges  usually  has  a  tensile 
strength  of  from  150  ooo  to  200  ooo  pounds  per  square  inch.  By 
compressing  steel  while  it  is  fluid,  the  strength  may  also  be  much 
increased,  and  this  process  is  used  for  the  steel  from  which  large 
guns  and  hollow  shafts  are  made. 

Annealing  consists  in  raising  cold  steel  to  a  light  red  heat 
and  then  allowing  it  to  cool  for  several  days.  This  process  reduces 
the  ultimate  strength,  but  it  increases  the  ductility.  As  an 
example,  the  following  table  gives  some  of  the  results  from  a 
large  series  of  specimens  prepared  by  the  Bethlehem  Steel  Com- 
pany in  1893  and  now  kept  at  Lehigh  University;  the  table 
refers  to  flat  bars  of  Bessemer  steel.  Art.  119  shows  that  anneal- 
ing increases  the  capacity  of  steel  to  resist  work. 

Tempering  consists  in  plunging  heated  steel  into  a  bath  of 
water  or  oil,  or  by  applying  these  fluids  to  its  surface.  The 
hardness  of  the  steel  and  its  ultimate  strength  are  thereby  much 
increased.  Armor  plate  undergoes  special  processes  of  temper- 


ART.  25 


STEEL. 


65 


ing  or  carbonization  which  render  it  excessively  hard  and  tough 
in  order  that  it  may  resist  projectiles  which  strike  it. 


Tensile  Strength 

Ultimate  Elongation 

Percent  of 

Pounds  per  Square  Inch 

Percentage 

Carbon 

Unannealed 

Annealed 

Unannealed 

Annealed 

0.08 

58  000 

56  ooo 

27 

31 

0.25 

84  ooo 

75000 

21 

25 

0.50 

125  ooo 

99  ooo 

II 

19 

0.67 

136  ooo 

112  OOO 

6 

16 

1  .04 

153  ooo 

128  ooo 

3* 

ii 

Steel  castings  are  extensively  used  for  axle  boxes,  cross-heads, 
and  machine  frames.  They  range  in  tensile  strength  from  60  ooo 
to  90  ooo  pounds  per  square  inch  and  have  an  elastic  limit  of 
somewhat  less  than  half  the  ultimate  strength.  Although  less 
reliable  than  steel  forgings,  they  give  excellent  service  after 
having  been  annealed  so  as  to  increase  their  ductility  and  their 
capacity  to  withstand  shock  and  work. 

Steel  is  alloyed  with  chromium,  manganese,  tungsten,  and  other 
materials  in  order  to  increase  its  strength,  hardness,  and  tough- 
ness. Modern  tool  steels  contain  from  five  to  ten  percent  of 
tungsten,  which  enables  the  tool  to  retain  its  temper  even  when 
under  a  red  heat.  Nickel  is  much  used  as  an  alloy  for  the  steel 
of  guns  and  armor  plates.  Nickel  steel  has  been  used  to  a  slight 
extent  for  structural  purposes  and  for  railroad  rails;  it  contains 
about  3^  percent  of  nickel,  and  has  an  elastic  limit  of  about 
48  ooo  and  a  tensile  strength  of  about  90  ooo  pounds  per  square 
inch.  Nickel  steel  has  been  made  with  an  elastic  limit  of  1 20  ooo 
and  a  tensile  strength  of  277  ooo  pounds  per  square  inch,  the 
ultimate  elongation  being  about  3  percent. 

Prob.  25a.  If  steel  costs  3  cents  per  pound  and  nickel  costs  35  cents 
per  pound,  what  should  be  the  cost  of  a  pound  of  nickel  steel  which 
contains  3.25  percent  of  nickel? 

Prob.  256.  Consult  Campbell's  paper  on  Tensile  Strength  of  Open- 
Hearth  Steel,  in  Transactions  of  American  Institute  of  Mining  Engi- 
neers, 1905,  and  test  his  Table  XVII  by  the  above  formulas. 


66  MATERIALS  OF  ENGINEER  NG  CHAP,  in 

ART.  26.     OTHER  MATERIALS 

Several  kinds  of  artificial  stone  have  been  made  since  1870, 
most  of  them  having  hydraulic  cement  and  sand  as  the  prin- 
cipal ingredients.  Beton  consists  only  of  those  materials  which 
are  subject  to  prolonged  trituration,  so  that  its  strength  is  much 
greater  than  ordinary  concrete.  Frear  stone  is  made  from  cement 
and  sand  with  a  small  quantity  of  gum  shellac.  Ransome  stone 
is  made  from  sand  and  sodium  silicate  which  are  thoroughly 
incorporated  in  molds  and  then  the  blocks  are  put  under  pres- 
sure in  a  hot  solution  of  calcium  chloride.  Sand-lime  brick, 
which  has  been  extensively  made  since  1900,  is  an  artificial  stone 
made  by  consolidating  heated  sand  and  lime  under  pressure. 
These  artificial  stones  are  used  mainly  for  the  walls  of  buildings, 
for  window  lintels,  and  for  steps. 

Ropes  are  made  of  hemp,  of  manilla,  and  of  iron  or  steel 
wire  with  a  hemp  center.  A  hemp  rope  one  inch  in  diameter 
has  an  ultimate  strength  of  about  6  ooo  pounds,  and  its  safe 
working  strength  is  about  800  pounds.  A  manilla  rope  is  slightly 
stronger.  Iron  and  steel  ropes  one  inch  in  diameter  have  ulti- 
mate strengths  of  about  36  ooo  and  50  ooo  pounds  respectively, 
the  safe  working  strengths  being  6  ooo  and  8  ooo  pounds.  As  a  fair 
rough  rule,  the  strength  of  ropes  may  be  said  to  vary  as  the  squares 
of  their  diameters,  that  is,  with  the  areas  of  the  cross-sections. 

Phosphor  bronze  is  an  alloy  of  copper  and  tin  containing 
from  2  to  6  percent  of  phosphorus.  It  is  remarkable  for  its 
complete  fluidity  so  that  most  perfect  castings  can  be  made. 
It  has  been  used  for  journal  bearings,  valve  seats;  and  even  for 
cannon.  It  is  hard  and  tough,  and  its  ultimate  tensile  strength 
may  range  from  40  ooo  to  100  ooo  pounds  per  square  inch. 

Aluminum  is  a  silver-gray  metal  which  is  malleable  and  duc- 
tile and  not  liable  to  corrode.  Its  specific  gravity  is  about  2.65, 
so  that  it  weighs  only  168  pounds  per  cubic  foot.  Its  ultimate 
tensile  strength  is  about  25  ooo  pounds  per  square  inch.  It  has 
a  low  modulus  of  elasticity,  and  its  ultimate  elongation  is  small. 
Alloys  of  aluminum  and  copper  have  been  made  with  a  tensile 


ART.  26  OTHER    MATERIALS  67 

strength  and  elongation  exceeding  those  of  wrought  iron,  but 
have  not  come  into  use  as  structural  materials. 

Numerous  brasses  and  bronzes  composed  of  copper,  tin, 
and  zinc  have  been  made.  The  strongest  of  these  alloys  was 
ascertained  by  Thurston  to  be  that  composed  of  55  parts  of  cop- 
per, 43  of  zinc,  and  2  of  tin,  its  ultimate  tensile  strength  being 
68  900  pounds  per  square  inch  with  an  elongation  of  48  per  cent 
and  a  reduction  of  area  of  70  percent. 

Brass,  which  is  composed  of  copper  and  zinc,  is  almost  the 
only  alloy  which  has  come  into  extensive  use  in  the  arts  and 
which  at  the  same  time  is  a  fully  reliable  material.  In  the  form 
of  castings  it  has  a  tensile  strength  of  about  20  ooo  pounds  per 
square  inch,  in  the  form  of  rolled  sheets  or  wire  it  has  a  much 
greater  strength.  Brass  water-pipes  are  now  frequently  used 
in  houses  by  those  who  can  afford  to  pay  as  high  a  price  as 
20  cents  per  pound. 

The  tensile  strength  of  lead  is  only  about  one-tenth  of  that  of 
brass,  and  it  attains  a  permanent  set  under  a  very  low  stress;  it  is 
indeed  almost  devoid  of  elasticity,  but  has  high  plasticity.  Glass 
is  very  brittle,  but  under  slowly  applied  loads  it  has  a  tensile 
strength  of  about  5  ooo  and  a  compressive  strength  of  about 
8  ooo  pounds  per  square  inch.  Anthracite  coal  is  a  brittle  mate- 
rial, having  a  compressive  strength  of  from  5  ooo  to  15  ooo  pounds 
per  square  inch. 

In  conclusion  it  may  be  noted  that  the  strength  and  other 
properties  of  materials  are  subject  to  variation  with  temperature, 
the  mean  values  given  in  the  preceding  pages  being  for  a  tem- 
perature ranging  from  40  to  90  degrees  Fahrenheit.  Experi- 
ments have  shown  that  wrought  iron  and  steel  continually  increase 
in  strength  under  static  loads  with  decreasing  temperatures,  the 
rate  of  increase  being  about  four  percent  for  each  100  degrees 
of  decrease  in  temperature.  This  rule  applies  for  all  ordinary 
temperatures  to  which  structures  and  machinery  are  subjected, 
and  to  temperatures  below  200  degrees  Fahrenheit.  As  the  tem- 
perature rises  above  200°,  wrought  iron  and  steel  increase  in 
strength,  attaining  a  maximum  at  about  500°  and  a  rapid  decrease 


68  MATERIALS  OF  ENGINEERING  CHAP,  ill 

then  follows  for  higher  temperatures;  the  elastic  limit,  however, 
seems  continually  to  decrease  as  the  temperature  rises.  Steel  and 
other  metals  are  more  brittle  at  low  temperatures  than  at  high 
ones,  and  the  breaking  of  railroad  rails  in  cold  weather  is  attrib- 
uted to  brittleness  and  not  to  decrease  in  static  strength. 

For  further  information  regarding  the  materials  discussed  in 
this  chapter  and  also  regarding  plaster,  leather,  india  rubber, 
glass,  ice,  and  glue,  see  American  Civil  Engineers'  Pocket  Book. 

For  information  concerning  methods  and  apparatus  for  testing 
materials,  see  the  same  Pocket  Book,  and  also  Chap.  XX  of  this 
volume. 

Prob.  260.  A  bar  of  aluminium  copper,  iJXif  inches  in  section 
area,  breaks  under  a  tension  of  42  800  pounds.  What  tension  will 
probably  break  a  bar  of  the  same  material  which  is  i|X2^  inches 
in  section  area? 

Prob.  266.  Compute  the  factor  of  safety  of  a  cast-iron  block,  2X2 
feet  in  section  area,  when  supporting  a  load  of  2  165  net  tons.  What 
should  be  the  size  of  a  white-oak  block  to  carry  the  same  load  with  a 
factor  of  safety  of  10? 


ART.  27  STRESS  DUE    TO    OWN   WEIGHT  69 

CHAPTER  IV 

CASES    OF    SIMPLE    STRESS 
ART.  27.    STRESS  DUE  TO  OWN  WEIGHT 

Tension,  compression,  and  shear  are  often  called  the  three 
cases  of  simple  stress,  it  being  understood  that  the  body  or  bar 
is  under  only  one  of  these  stresses  at  the  same  time.  In  tension 
and  compression  the  loads  act  along  the  axis  of  the  bar,  in  shear 
they  act  normally  to  it.  When  a  bar  is  vertical,  its  weight  causes 
stresses  additional  to  those  arising  from  the  applied  load,  and 
these  will  now  be  considered. 

When  an  unloaded  bar  is  hung  vertically  by  one  end,  there 
is  no  stress  on  the  lower  end,  while  at  the  upper  end  there  is 
a  tensile  stress  equal  to  the  weight  of  the  bar.  When  a  vertical 
bar  of  weight  W  has  a  load  P  at  its  lower  end,  the  tensile  stress 
at  that  end  is  P  and  that  at  the  upper  end  is  P+W.  In  many 
practical  cases  W  is  small  compared  with  'P,  so  that  it  is  unneces- 
sary to  consider  it;  a  common  rule  is  that  the  stress  due  to  W 
need  not  be  regarded  when  it  is  less  than  ten  percent  of  that 
due  to  P.  A  steel  bar  one  square  inch  in  section  area  and  30  feet 
long  weighs  about  102  pounds  (Art.  17),  and  a  stress  of  100  pounds 
per  square  inch  is  small  compared  with  the  elastic  limit  of  the 
material. 

The  limiting  length  of  an  unloaded  vertical  bar  is  that  at 
which  it  would  break  at  the  upper  end  under  its  own  weight. 
Let  a  be  the  section  area  of  the  bar,  S  the  ultimate  strength  of 
the  material,  v  its  weight  per  cubic  unit,  and  /  the  limiting  length. 
Then  the  weight  of  the  bar  is  val,  the  tensile  stress  at  its  upper 
end  is  aS,  and  hence  for  rupture  S  =  vl  , whence  /  =S/v.  For 
example,  take  a  cast-iron  bar  for  which  S  =20000  pounds  per 
square  inch  or  2  880  ooo  pounds  per  square  foot,  and  v  =450 
pounds  per  cubic  foot;  then  the  limiting  length  is  /  = 
2880000/450=6400  feet.  Of  course,  no  vertical  bar  of  this 
length  is  possible. 


70 


CASES  OF  SIMPLE  STRESS 


CHAP.  IV 


The  limiting  length  of  a  vertical  bar  loaded  at  its  lower  end 
is  that  at  which  it  would  break  at  the  upper  end  under  the  stress 
due  to  the  load  and  its  own  weight.  If  P  is  the  load  at  the  lower 
end,  the  stress  at  the  upper  end  is  P  +  val,  and  hence  for  rupture 
Sa=P  +  val,  from  which  l  =  (S-P/a)/v.  Hence  P/a  is  the 
unit-stress  due  to  P,  and  the  formula  shows  that  /  is  zero  when 
P/a  equals  the  ultimate  tensile  strength  of  the  material.  For 
a  cast-iron  bar  let  P/a  be  10  ooo  pounds  per  square  inch,  then 
the  limiting  length  is  3  200  feet. 


Fig.  27a  Fig.  27b 

The  elongation  of  a  vertical  bar  under  its  own  weight  is  one- 
half  of  that  caused  by  the  same  load  applied  at  the  lower  end. 
To  show  this,  let  /  be  the  length,  a  the  section  area,  and  W  the 
weight  of  the  bar.  Let  oc  be  any  distance  from  the  lower  end 
(Fig.  27<z),  then  W.  x/l  is  the  weight  of  this  portion.  The  ele- 
mentary elongation  caused  by  this  weight  on  the  elementary 
length  dx  is  from  (10)  given  by  W(x/l)dx/aE,  where  E  is  the 
modulus  of  elasticity.  The  integral  of  this  between  the  limits 
I  and  0  gives  e  =  %Wl/aE,  which  is  one-half  the  elongation  due 
to  a  load  W  at  the  end  (Art.  10).  In  a  similar  manner  the  elonga- 
tion due  to  the  weight  of  the  bar  and  a  load  P  at  the  end  is  found 
to  be  e  =  (%W  +  P)l/aE,  and  this  is  the  sum  of  the  separate  elonga- 
tions due  to  W  and  P.  These  expressions  apply  also  to  the 
corresponding  cases  of  compression  shown  in  Fig.  276,  but  here, 
as  in  tension,  the  formulas  apply  only  when  the  greatest  unit- 
stress  does  not  exceed  the  elastic  limit  of  the  material. 

Prob.  27a.  Find  the  length  of  a  vertical  wooden  bar,  6X6  inches  in 
section  area  and  having  a  load  of  21  600  pounds  at  the  lower  end,  so 
that  the  stress  at  the  upper  end  shall  be  650  pounds  per  square  inch. 

Prob.  27b.  Compute  the  elongation  of  the  above  bar  due  to  the 
load  at  the  end  and  that  due  to  its  own  weight. 


ART.  28  BAR  OF  UNIFORM  STRENGTH  71 

ART.  28.    BAR  OF  UNIFORM  STRENGTH 

A  suspension  rod  of  constant  cross-section  is  stressed  at  the 
lower  end  by  the  load  P,  and  at  the  upper  end  by  P  plus  the 
weight  of  the  rod.  When  the  rod  is  very  long  its  section  area 
should  be  less  at  the  lower  than  at  the  upper  end  in  order  both 
to  economize  material  and  to  reduce  its  weight.  The  rod  in 
such  cases  is  sometimes  made  in  parts,  the  section  area  of  any 
part  being  less  than  that  of  the  one  above  it. 

A  vertical  tension  bar  of  uniform  strength  is  one  in  which 
the  unit-stress  is  the  same  in  all  section  areas.  The  theoretic 
form  for  such  a  bar  will  now  be  de-  - 
termined.  Let  P  be  the  load  applied 
at  the  lower  end,  and  S  the  allow- 
able unit-stress;  then  the  section 
area  of  the  lower  end  is  aQ  =  P/Sm 

Let  a  be  the  section  area  at  a   dis-  v.     oc 

rig.  28 

tance  y  from  the  lower  end;    then 

a+da  is  the  area  at  the  distance  y  +  dy,  and  the  area  da  must 
be  sufficient  to  resist  the  stress  due  to  the  weight  of  the  bar  in 
the  distance  dy.  Let  v  be  the  weight  of  a  cubic  unit  of  the  mate- 
rial, then  the  weight  of  the  bar  in  the  distance  dy  is  va  .  dy,  and 
hence  da  =  (va .  dy)/S,  which  may  be  written  dy  =  (S/v)(da/a) 
Integrating  this  and  determining  the  constant  by  the  condition 
that  a  equals  a0  when  y  is  o,  there  is  found, 

y=  (S/v)  (log*a-  log^o) 

in  which  the  logarithms  are  in  the  Naperian  system.  Passing 
to  common  logarithms  by  the  well-known  rules,  this  becomes, 

log  a = 0.4343  (v/S)y+ log  a0 

which  is  the  formula  for  computing  a  at  any  distance  y  from  the 
lower  end,  a0  being  first  found  from  a0=P/S.  This  formula 
applies  to  compression  as  well  as  to  tension,  and  the  second  dia- 
gram in  Fig.  28  illustrates  this  case. 

For  example,  take  a  round  masonry  pier  which  is  to  carry 
a  load  of  900  ooo  pounds  with  an  allowable  working  stress  of 
100  pounds  per  square  inch  or  14  400  pounds  per  square  foot. 


72  CASES  OF  SIMPLE  STRESS  CHAP.  IV 

The   area  of   the   top  is   #o  =900/14.4  =  62.5    square   feet,    and 
v/S  =  160/14  400  =0.01 1 1 1.     The  formula  then  becomes  log  a  = 
0.00482^+1.7959.      The    following    values    of   the    section    area 
y  =  o         10  20  30  40  feet 

a  =  62. 5      69.8        78.0          87.2          97.4  square  feet 
d=  8.92       9.43         9.97         10.54         ii. 14  feet 

and  the  diameter  of  the  pier  are  now  computed  for  different 
values  of  the  height  y,  and  it  is  seen  that  the  profile  of  the  pier 
is  slightly  curved.  The  extra  expense  of  construction  of  a  pier 
of  uniform  strength  is,  however,  usually  greater  than  that  of  the 
extra  amount  of  material  of  a  trapezoidal  profile,  so  that  the 
latter  is  generally  used.  A  round  pier  with  trapezoidal  profile 
carrying  the  same  load  as  above  with  a  working  stress  of  100 
pounds  per  square  inch  on  top  and  base,  requires  a  section  area 
of  97.8  square  feet  at  the  base. 

Prob.  28.  A  vertical  steel  rod  of  a  mine  pump  which  is  to  carry  a 
load  of  40  ooo  pounds  at  its  lower  end,  is  required  to  have  no  stress 
greater  than  6  ooo  pounds  per  square  inch,  its  length  being  185  feet. 
Compute  its  section  area  if  it  is  of  uniform  size.  Compute  the  section 
area  at  the  upper  end  if  it  could  be  made  of  uniform  strength. 

ART.  29.     ECCENTRIC  LOADS 

In  all  the  discussions  of  the  preceding  articles  the  resultant 
of  the  tensile  or  compressive  load  on  a  bar  has  been  supposed 
to  coincide  in  direction  with  the  axis  of  the  bar  and  the  unit- 
stress  produced  by  it  to  be  uniformly  distributed  over  each  sec- 
tion area  (Art.  1).  This  is  the  common  case  of  simple  axial 
stress,  but  there  are  also  cases  where  the  load  is  ' eccentric',  that 
is,  it  does  not  coincide  with  the  axis;  the  effect  of  this  is  to  cause 
the  unit-stress  to  be  greater  on  one  side  of  the  section  than  on 
the  other.  If  P  is  the  load  on  a  bar  and  a  is  its  section  area, 
then  for  axial  loads  P/a  is  the  actual  uniform  unit-stress;  for 
eccentric  loads  P/a  is  a  mean  or  average  value,  some  of  the  unit- 
stresses  being  greater  and  others  being  less  than  P/a. 

Fig.  29a  shows  part  of  a  rectangular  bar  where  the  eccentric 
tensile  load  P  is  applied  at  the  distance  p  from  the  axis  of  the 


ART.  29  ECCENTRIC   LOADS  73 

bar  and  in  the  middle  of  the  width  of  the  section,  and  Fig.  296 
shows  the  same  bar  under  compression.  At  any  section  mn 
the  resisting  unit-stresses  act  in  the  opposite  direction  to  P  and 
hold  it  in  equilibrium.  The  resultant  of  these  unit-stresses  must 
equal  the  load  P  in  order  that  there  may  be  no  motion  in  the 
direction  of  the  length  of  the  bar,  and  it  must  be  in  the  same  line 
of  action  as  P  in  order  that  there  may  be  no  motion  of  rotation. 
From  these  two  conditions  of  equilibrium  (Elements  of  Mechanics, 
Art.  5)  the  values  of  the  greatest  unit-stress  Si  and  the  least 
unit-stress  £2  can  be  found  when  the  law  of  variation  of  the 
stresses  across  the  section  is  known. 


Fig.  29a  Fig.  296 

It  has  been  ascertained  by  experiment  that,  when  the  elastic 
limit  of  the  material  is  not  exceeded  the  stresses  increase  across 
the  section  at  a  uniform  rate  from  S2  to  Si,  and  hence  the  law 
of  variation  is  that  of  a  straight  line.  From  this  law,  the  mean 
unit-stress  is  i(Si  +  S2)  and  the  total  stress  on  the  section  area 
a  is  i(Si+S2)a;  accordingly  the  first  condition  of  equilibrium 
is  expressed  by  J(Si  +  S2)a  =  P.  To  express  the  second  con- 
dition, it  may  be  noted  that  the  resultant  load  P  must  be  oppo- 
site to  the  center  of  gravity  of  the  trapezoid  which  represents 
the  unit-stresses;  or  conversely,  the  center  of  gravity  of  this 
trapezoid  must  be  at  the  distance  p  above  the  axis  of  the  bar 
in  Fig.  29a  and  at  the  distance  p  below  the  axis  in  Fig.  296.  Let  d 
be  the  width  of  the  rectangular  section;  then  it  is  known  that 
the  center  of  gravity  of  the  trapezoid  having  the  two  parallel 
sides  Si  and  S2  and  the  altitude  d  is  given  by  p  = 
±d(Si  -S2)/(Si  +S2).  Solving  these  two  equations  there  results, 


from  which  the  unit-stresses  may  be  computed  for  any  given  value 
of  the  eccentricity  p.  When  p=o,  both  Si  and  S2  are  equal 
to  P/a  and  the  stress  is  uniformly  distributed. 


74 


CASES  OF  SIMPLE  STRESS 


CHAP.  IV 


The  following  figures  show  the  distribution  of  the  unit-stresses 
for  several  cases  of  compression,  the  cross-section  of  each  prism 
being  rectangular  and  having  the  width  d.  In  Fig.  29c  the  load 
is  axial  or  p=o  and  Si=S2=P/a-  In  Fig.  29d  the  distance 
p  is  less  than  \d  and  the  values  of  S\  and  £2  are  given  by  the 
above  formulas.  In  Fig.  29e  the  distance  p  is  \d  and  the  formulas 
give  Si=2P/a  and  £2=0,  so  that  the  unit-stress  on  the  side  of 
the  section  nearest  P  is  double  that  of  uniform  distribution,  while 
there  is  no  stress  on  the  other  side.  In  Fig.  29/  the  distance  / 
is  greater  than  %d,  so  that  Si  is  greater  than  2P/a,  while  ,$2  is 
negative,  that  is,  6*2  is  tension  instead  of  compression.  As  a 
lumerical  example,  let  a  masonry  pier  having  a  base  10X30  feet, 
as  in  Fig.  29g,  carry  a  load  of  i  200  ooo  pounds  on  its  top  at  a 
distance  of  6  feet  from  the  middle.  Here  P/a  =  i  200  000/300  = 
4  ooo  pounds  per  square  foot  =  28  pounds  per  square  inch,  and 
p/d  =  6/30  =0.2.  Accordingly  the  greatest  compressive  stress 
due  to  this  eccentric  load  is  Si  =28(1  +  1.2)  =62  pounds  per 
square  inch,  while  the  stress  on  the  other  side  is  $2  =  28(1  — 1.2)  = 
about  6  pounds  per  square  inch  tension. 

ip 


fflUJ 


s, 


Utn 


w\ 


-  d-  - 


Fig.  29c  Fig.  29d        Fig.  29e  Fig.  29/ 


Fig 


The  above  investigation  considers  only  the  stresses  caused  by 
the  eccentric  load  and  takes  no  account  of  those  due  to  the  weight 
of  the  bar  or  prism  itself.  If  this  weight  be  W  the  uniform 
unit-stress  due  to  it  is  W/a,  and  this  may  be  added  to  those  caused 
by  the  eccentric  load  P.  For  instance,  let  the  pier  in  the  above 
example  weigh  1500000  pounds;  then  W/a  =  35  pounds  pel 
square  inch,  so  that  the  greatest  and  least  compressive  stresses 
due  to  both  P  and  W  are  5*1=35  +  62=97  and  52  =  35—6=29 
pounds  per  square  inch. 

When  the  load  on  a  rectangular  prism  has  no  eccentricity, 
the  stress  on  each  section  is  uniformly  distributed  and  hence  all 


ART.  30  WATER  AND  STEAM  PIPES.  75 

parts  of  the  prism  suffer  the  same  change  of  length.  For  an 
eccentric  load,  the  stresses  vary  throughout  the  section,  and 
hence  the  changes  in  length  are  not  uniform;  thus  in  Fig.  29e 
the  left-hand  side  shortens  the  amount  Si/E  for  every  unit  of 
length  (Art.  10),  while  the  right-hand  side  suffers  no  change  of 
length;  again  in  Fig.  29/  the  left-hand  side  shortens  but  the 
right-hand  side  elongates.  The  result  of  these  unequal  changes 
of  length  is  to  cause  the  prism  or  bar  to  bend  laterally,  and  this 
lateral  deflection  is  discussed  in  Arts.  86-87,  where  also  formulas 
for  cross-sectiors  of  any  shape  are  deduced. 

Prob.  29a.  A  rectangular  wooden  block  is  6X16  inches  in  section 
area  and  18  inches  high.  It  carries  two  loads  applied  on  the  top  at 
the  middle  of  the  width  of  the  cross-section,  PI  being  2  inches  from 
the  center,  while  P2  is  3  inches  from  the  center  but  on  the  side  oppo- 
site PI.  Compute  the  ratio  of  PI  to  PZ  so  that  the  unit-stress  may  be 
uniform  over  the  base. 


ART.  30.    WATER  AND  STEAM  PIPES. 

The  pressure  of  water  or  steam  in  a  pipe  is  exerted  in  every 
direction  as  shown  in  the  transverse  section  of  Fig.  300;  this 
tends  to  tear  the  pipe  apart  longi- 
tudinally and  it  is  resisted  by  the 
tensile  stresses  of  the  material. 
Let  R  be  the  pressure  per  unit  of 
area  which  is  exerted  by  the  water 
or  steam,  d  the  diameter  of  the  pipe,  Fig-  30a  F[%-  m 

and  /  its  length.  Then  the  force  which  tends  to  cause  longitudinal 
rupture  is  Id .  R;  this  follows  from  the  principle  of  hydrostatics 
that  the  pressure  of  a  fluid  or  gas  in  any  direction  is  equal  to  the 
pressure  on  a  plane  normal  to  that  direction,  or  it  may  be  shown 
by  imagining  the  pipe  to  be  filled  with  a  solid  substance  on  one 
side  of  the  diameter,  as  in  Fig.  306,  which  receives  the  pressure  R 
on  each  unit  of  the  area  Id.  Let  /  be  the  thickness  of  the  pipe 
and  S  the  resisting  tensile  unit-stress;  when  /  is  small  compared 
with  d  the  tensile  stress  may  be  regarded  as  uniformly  distributed 
and  then  the  total  resisting  stress  is  2# .  S.  Since  the  resisting 


76  CASES  OF  SIMPLE  STRESS  CHAP.  IV 

stress  must  equal  the  acting  pressure,  it  follows  that, 

2tS=dR  or  S/R  =  d/2t  (30) 

which  is  the  common  formula  for  pipes  under  internal  pressure. 

The  unit-pressure  R  for  water  may  be  computed  from  a  given 
head  h  by  finding  the  weight  of  a  column  of  water  of  that  height 
and  one  square  unit  in  section.  Or,  if  h  be  given  in  feet,  the 
unit-pressure  in  pounds  per  square  inch  may  be  computed  from 
7?=o.434/t  (Treatise  on  Hydraulics,  Art.  11). 

Pipes  are  made  of  cast  iron,  wrought  iron,  and  steel.  Cast 
iron  is  used  for  water  pipes  up  to  48  inches  in  diameter,  and 
steel  is  generally  used  for  steam  pipes.  Large  water  pipes  are 
made  of  steel  plates  riveted  together,  but  the  discussion  of  these 
is  reserved  for  another  article.  A  water  pipe  subject  to  the  shock 
of  water  ram  requires  a  high  factor  of  safety,  and  in  a  steam 
pipe  the  factors  should  also  be  high  owing  to  the  shocks  liable 
to  occur  from  the  condensation  and  expansion  of  the  steam. 
The  above  formula  shows  that  the  thicknesses  of  thin  pipes  of 
the  same  material  under  the  same  internal  pressure  should  increase 
directly  as  their  diameters.  Thick  pipes  are  treated  in  Art.  150. 

For  example,  let  it  be  required  to  find  the  factor  of  safety 
of  a  cast-iron  water  pipe  of  12  inches  diameter  and  f  inches 
thickness  under  a  head  of  300  feet.  Here  R  is  0.434X300  =  130.2 
pounds  per  square  inch.  Then  the  unit-stress  is 

,S=i2Xi3o.2/(2Xf)  =  i  250  pounds  per  square  inch 

and  hence  the  factor  of  safety  of  the  cast  iron  under  this  tensile 
stress  is  20000/1  250=  about  16,  which  indicates  ample  security 
under  ordinary  conditions. 

Again,  let  it  be  required  to  find  the  proper  thickness  for  a 
wrought-iron  steam  pipe  of  18  inches  diameter  to  resist  a  pres- 
sure of  120  pounds  per  square  inch.  With  a  factor  of  safety 
of  10  the  working  unit-stress  S  is  about  5  ooo  pounds  per  square 
inch.  Then  from  the  formula  the  required  thickness  is  /  = 
(120/5  ooo) XQ  =0.22  inches.  In  order  safely  to  resist  the  shocks 
liable  to  occur  in  handling  the  pipes,  the  thickness  is  often  made 
greater  than  the  above  formula  requires. 


ART.  31  THIN  CYLINDERS  AND  SPHERES  77 

Prob.  30a.  What  should  be  the  thickness  of  a  cast-iron  water  pipe 
of  1 8  inches  diameter  under  a  head  of  300  feet,  the  factor  of  safety 
being  taken  as  15? 

Prob.  306.  A  wrought-iron  pipe  is  3  inches  in  internal  diameter  and 
weighs  8  pounds  per  linear  foot.  Compute  its  thickness,  and  the  pres- 
sure it  can  carry  with  a  factor  of  safety  of  10. 

ART.  31.     THIN  CYLINDERS  AND  SPHERES 

A  cylinder  subject  to  the  interior  pressure  of  water  or  steam 
tends  to  fail  longitudinally  exactly  like  a  pipe.  The  head  of 
the  cylinder,  however,  undergoes  a  pressure  which  tends  to 
separate  it  from  the  walls.  If  d  is  the  diameter  of  the  cylinder 
and  R  the  internal  pressure  per  square  unit,  the  total  pressure 
on  the  head  is  jrcd2  .  R.  If  S  is  the  working  unit-stress  and  / 
the  thickness  of  the  cylinder,  the  resistance  to  the  pressure  is 
approximately  ndt .  S  when  /  is  so  small  that  S  may  be  regarded 
as  uniformly  distributed.  Since  the  resisting  stress  must  equal 
the  acting  pressure, 

ndt.S=\nd2  .R     or    S/R=d/tf 

By  comparing  this  with  the  formula  of  the  last  article  it  is  seen 
that  the  resistance  of  a  pipe  to  transverse  rupture  is  double  the 
resistance  to  longitudinal  rupture. 

A  thin  sphere  subject  to  interior  pressure  tends  to  rupture 
around  a  great  circle,  and  it  is  easy  to  see  that  the  conditions 
are  exactly  the  same  as  for  the  transverse  rupture  of  a  cylinder, 
or  that  4tS=dR.  For  thick  spheres  and  cylinders  the  formulas 
of  this  and  the  last  article  are  only  approximate;  a  full  discus- 
sion of  these  will  be  found  in  Arts.  150,  152,  163. 

A  cylinder  under  exterior  pressure  is  theoretically  in  a  simi- 
lar condition  to  one  under  interior  pressure  as  long  as  it  remains 
a  true  circle  in  cross-section.  A  uniform  interior  pressure  tends 
to  preserve  and  maintain  the  circular  form  of  the  cylindrical 
annulus,  but  an  exterior  pressure  tends  at  once  to  increase  the 
slightest  variation  from  the  circle  and  render  it  elliptical.  The 
distortion  when  once  begun  rapidly  increases,  and  failure  occurs 
by  the  collapsing  of  the  tube  rather  than  by  the  crushing  of  the 


78  CASES  OF  SIMPLE  STRESS  CHAP.  IV 

material.  The  flues  of  a  steam  boiler  are  the  most  common 
instance  of  cylinders  subjected  to  exterior  pressure.  In  the 
absence  of  a  rational  method  of  investigating  such  cases  recourse 
has  been  had  to  experiment.  Tubes  of  various  diameters,  lengths, 
and  thicknesses  have  been  subjected  to  exterior  pressure  until 
they  collapse  and  the  results  have  been  compared  and  discussed. 
The  following,  for  instance,  are  the  results  of  three  experiments 
by  Fairbairn  on  wrought-iron  tubes,  the  collapsing  pressure 
being  in  pounds  per  square  inch: 

Length  37  60  61       inches 

Diameter  9  14  J  i8|     inches 

Thickness  0.14  0.125  0.25  inches 

Unit-pressure  378  125  420 

Frorr  these  and  other  similar  experiments  it  has  been  concluded 
that  the  collapsing  unit-pressure  varies  directly  as  some  power 
of  the  thickness,  and  inversely  as  the  length  and  diameter  of 
the  tube.  For  wrought-iron  tubes  Wood  gives  the  empirical 
formula  for  the  collapsing  pressure  per  square  inch, 

R=96oooool2'18/ld 

and  the  values  of  R  computed  from  this  formula  for  the  above 
three  experiments  are  397,  120,  and  409  pounds  per  square  inch 
which  agree  well  with  the  observed  values. 

The  proper  thickness  of  a  wrought-iron  tube  to  resist  exterior1 
pressure  may  be  readily  found  from  this  formula  after  assuming 
a  suitable  factor  of  safety.  For  example,  let  it  be  required  to 
find  /  when  R  =  120  pounds  per  square  inch,  1  =  J2  inches,  .d  =  4 
inches,  and  the  factor  of  safety  =  10.  Then  the  formula  gives 
/2*18=  0.036  from  which,  with  the  help  of  logarithms,  the  thick- 
ness /  is  found  to  be  0.22  inches. 

Prob.  3 la.  What  interior  pressure  per  square  inch  will  burst  a  cast- 
iron  sphere  of  24  inches  diameter  and  j  inches  thickness  ? 

Prob.  316.  What  exterior  pressure  per  square  inch  will  collapse  a 
wrought-iron  tube  72  inches  long,  4  inches  diameter,  and  0.25  inches 
thickness?  What  is  a  proper  thickness  for  this  tube  under  a  steam 
pressure  of  1 50  pounds  per  square  inch  ? 


ART.  32  SHRINKAGE    OF    HOOPS  79 

ART.  32.     SHRINKAGE  OF  HOOPS 

Hoops  and  tires  are  frequently  turned  with  the  interior  diam- 
eter slightly  less  than  that  of  the  wheels  or  cylinders  upon  which 
they  are  to  be  placed.  A  hoop  is  expanded  by  heat  and  placed 
in  position,  and  in  cooling  it  shrinks  and  is  held  firmly  upon 
the  cylinder  by  the  radial  pressure  caused  by  the  shrinkage.  The 
effect  of  this  radial  pressure  is  to  cause  tension  in  the  hoop,  and 
compression  throughout  the  mass  that  it  encircles. 

When  the  hoop  is  thin  compared  to  the  diameter  of  the  cylinder 
upon  which  it  is  shrunk,  the  entire  deformation  due  to  the  shrink- 
age may  be  practically  regarded  as  confined  to  the  hoop.  The 
tangential  unit-stress  in  the  hoop  will  then  be  due  only  to  the 
increase  in  length  of  the  circumference,  and  this  will  be  pro- 
portional to  the  increase  in  its  diameter.  If  s  is  the  unit-elonga- 
tion of  the  hoop  and  E  the  modulus  of  elasticity  of  the  material, 
then  by  (9)  the  tensile  unit-stress  due  to  the  elongation  is  S  =  eE. 

Let  d  be  the  diameter  of  the  cylinder  upon  which  the  hoop 
is  to  be  shrunk,  and  d\  be  the  interior  diameter  to  which  the 
hoop  is  turned.  Supposing  that  d  is  unchanged  by  the  shrink- 
age, d\  will  be  increased  to  d,  and  the  unit-elongation  of  the 
inner  circumference  of  the  hoop  will  be  e  =  (d  —  di)/di'}  since  the 
hoop  is  thin,  this  is  practically  the  unit-elongation  for  all  parts 
of  the  hoop.  Accordingly, 


or 


is  the  tensile  unit-stress  in  the  hoop  due  to  shrinkage. 

A  common  rule  for  the  case  of  steel  hoop  shrinkage  is  to 
make  d  —  di  equal  to  T?Vo^>  that  is  the  hoop  is  turned  so  that  its 
interior  diameter  is  TT7nrtri  IGSS  than  the  diameter  of  the  cylinder; 
then  di=}$%$d,  and  e  =  (d  —  di)/di  =TTV?>  which  is  practically 
the  same  as  rgVir-  The  tangential  tensile  unit-stress  in  the 
hoop  then  is  ,5=30000000/1500  =  20000  pounds  per  square 
inch. 

The  radial  unit-stress  acting  between  the  cylinder  and  the 
hoop  is  from  formula  (30)  readily  found  to  be  R  =  2tS/dil  and 


80  CASES  OF  SIMPLE  STRESS  CHAP.  IV 

hence  its  value  depends  upon  the  diameter  of  the  cylinder  and 
the  thickness  of  the  hoop.  For  a  locomotive  driving  wheel 
having  d=6o  inches,  /  =  J  inches,  and  S  =20000  pounds  per 
square  inch,  the  radial  unit-pressure  R  between  the  tire  and  wheel 
is  500  pounds  per  square  inch. 

The  above  discussion  gives  values  of  S  somewhat  too  high, 
because  the  radial  pressure  acting  between  the  hoop  and  the 
cylinder  produces  some  decrease  in  the  diameter  of  the  latter. 
For  thick  hoops  upon  hollow  cylinders,  such  as  those  of  large 
guns,  the  above  method  is  not  sufficiently  accurate,  and  a  more 
exact  one  is  given  in  Art.  152. 

Prob.  32.  Upon  a  cylinder  18  inches  in  diameter  a  wrought-iron 
hoop  2  inches  thick  is  to  be  placed.  The  hoop  is  turned  to  an  interior 
diameter  of  17.98  inches  and  shrunk  on.  Compute  the  tensile  unit- 
stress  in  the  hoop. 

ART.  33.  INVESTIGATION  OF  RIVETED  JOINTS 

When  two  overlapping  plates  are  fastened  together  with  one 
row  of  rivets,  as  in  Fig.  33a,  the  joint  is  called  a  lap  joint  with 
single  riveting;  when  two  rows  are  used,  as  in  Fig.  336,  it  is 
said  to  be  a  lap  joint  with  double  riveting.  When  tension  is 
transmitted  through  such  plates,  its  first  effect  is  to  bring  a  side- 
wise  compression  on  the  rivet,  and  this  in  turn  brings  a  shear 
on  the  rivet  which  tends  to  cut  it  off  in  the  plane  of  the  surface 
of  junction  of  the  plate.  The  exact  manner  in  which  the  side- 
wise  compression  acts  upon  the  cylindrical  surface  of  the  rivet 
is  not  known,  but  it  is  usually  supposed  that  it  causes  a  com- 
pressive  stress  which  is  uniformly  distributed,  over  the  projec- 
tion of  that  surface  upon  a  plane  through  the  axis  of  the  rivet. 

For  a  lap  joint  with  single  riveting,  as  in  Fig.  33a,  let  P  be 
the  tensile  force  which  is  transmitted  from  one  plate  to  another 
by  means  of  a  single  rivet,  t  the  thickness  of  the  plate,  and  p 
the  pitch  of  the  rivets.  Let  St  be  the  tensile  unit-stress  which 
occurs  in  the  section  of  the  plate  between  two  rivets,  and  Sc  and 
Ss  be  the  unit-stresses  of  compression  and  shear  upon  a  rivet. 
Then  the  equations  between  the  stresses  and  the  force  P  are, 


ART.  33 


INVESTIGATION  OF  RIVETED  JOINTS 


81 


for  tension  on  plate,  t(p  —  d)St  =  P 

for  compression  on  rivet,  tdSc  =  P 

for  shear  on  rivet,  ±nd2  .  SS  =  P 

From  these  equations  the  unit-stresses  may  be  computed  when 
the  other  quantities  are  known,  and  by  comparing  them  with 
the  proper  allowable  unit-stresses  (Art.  7)  the  degree  of  security 
of  the  joint  is  estimated. 


FIG.  33a. 


FIG.  33Z>. 


For  a  lap  joint  with  double  riveting,  the  plates  have  a  wider 
lap,  and  the  two  rows  of  rivets  are  staggered,  as  in  Fig.  336. 
Let  P  be  the  tension  which  is  exerted  over  the  width  equivalent 
to  the  pitch  p,  this  being  the  distance  between  the  centers  of 
two  rivets  in  one  row.  Then  P  is  transferred  from  one  plate 
to  another  through  two  rivets,  and  the  three  formulas  are, 

t(p-d}St  =  P  2td.Sc  =  P  2.&d2.Ss  =  P 

from  which  the  unit-stresses  due  to  the  tension  P  may  be 
computed  and  the  security  of  the  joint  be  investigated.  This 
joint  is  usually  a  stronger  one  than  that  with  single  riveting,  if 
proper  values  are  assigned  to  p  and  d. 

The  investigation  of  a  given  riveted  joint  consists  in  deter- 
mining the  values  of  StJ  Ss,  and  Sc  from  the  above  equations 
and  then  computing  the  factors  of  safety  (Art.  7).  For  example 
take  a  lap  joint  with  double  riveting  where  P  =  8  ooo  pounds, 
p=2%  inches,  d  =  %  inches,  and  /  =  |  inches,  Then  the  tensile 
stress  on  the  plate  between  two  rivets  is, 

St  =  P/t(p  —  d)  =  6  560  pounds  per  square  inch. 


82  CASES  OF  SIMPLE  STRESS  CHAP.  IV 

while  the  sidewise  compressive  stress  on  the  rivet  is, 
Sc=P/2td=6  100  pounds  per  square  inch, 

and  the  shearing  stress  on  the  rivet  is, 

Ss=P/?nd2=6  650  pounds  per  square  inch. 

Now,  if  these  plates  and  rivets  be  of  structural  steel  having  an 
ultimate  tensile  and  compressive  strength  of  55  ooo  pounds  per 
square  inch  and  an  ultimate  shearing  strength  of  45  ooo  pounds 
per  square  inch,  the  factors  of  safety  are  8.4  for  the  plate  in  ten- 
sion, 9.0  for  the  rivet  in  compression,  and  6.8  for  the  rivet  in 
shear. 

The  '  efficiency  '  of  a  joint  is  defined  as  the  ratio  of  its  high- 
est allowable  stress  to  the  highest  allowable  stress  of  the  unriveted 
plate.  For  any  riveted  joint  three  efficiencies  may  be  computed 
by  dividing  the  three  values  of  P  by  pt .  St)  which  is  the  allow- 
able stress  on  the  unriveted  plate,  and  the  smallest  of  these  is 
the  efficiency  of  the  joint.  For  example,  let  St=S^=  55000, 
and  £5=45000  pounds  per  square  inch,  and  let  p  =  2^,  d=f, 
and  /=|  inches  for  a  lap  joint  with  double  riveting.  Then  for 
the  plate  in  tension  the  efficiency  is  (p  —  d)tSi/ptSt  =  (p  —  d)/p  = 
0.65;  for  the  rivet  in  compression  the  efficiency  is  2tdSc/ptSt  = 
2dSc/pS ',=0.70;  and  for  the  rivet  in  shear  the  efficiency  is 
%7id2Sg/ptSt=o.$2.  The  efficiency  of  this  joint  is  therefore  0.52, 
that  is,  its  strength  is  52  percent  of  that  of  an  unriveted  plate. 
In  a  properly  designed  joint  all  parts  are  of  equal  strength  and 
the  three  efficiencies  are  equal. 

A  butt  joint  is  one  in  which  there  is  no  overlapping  of  the 
main  plates,  but  cover  plates  are  used.  Fig.  33c  shows  a  case 

^-^ /x — ^  where  there  are  two  cover 

plates  and  a  single  row  of 

\ X  V — ^  rivets  on  each  side  of  the 

Flg'  33c  joint.  When  tension  is  ap- 

plied to  the  two  main  plates,  it  first  produces  compression  on 
the  rivets  and  this  in  turn  brings  the  rivets  into  shear.  This 
shear  comes  on  two  cross-sections  of  each  rivet  and  transfers 
one-half  of  the  applied  tension  into  each  cover  plate.  Accord- 
ingly the  thickness  of  a  cover  plate  should  be  one-half  of  that 


ART.  34  DESIGN    OF    RlVETED    JOINTS  83 

of  the  main  plate.     Let  the  notation  be  the  same  as  before;  then, 

P=t(p-  d)St  P=  tdSc  P=2.  \n&  .  S8 

are  the  three  formulas  for  the  cases  of  tension,  compression,  and 
shear.  Here  the  expressions  for  tension  and  compression  are 
the  same  as  those  for  a  lap  joint  with  single  riveting,  but  that  for 
shearing  is  different  because  the  total  stress  is  divided  between 
two  cross-sections  of  a  rivet. 

Butt  joints  having  two  cover  plates,  and  two  rows  of  rivets 
on  each  side,  are  also  used.  For  this  case  it  is  easy  to  see  that 
the  three  formulas  are, 

t(p-d)St  =  P  2.td.Sc=P  4.±nd2S8=P 

since  the  force  P  brings  compression  upon  two  rivets  and  shear 
upon  four  rivet  sections.  Here,  as  always,  /  is  the  thickness  of 
the  main  plates  and  p  is  the  pitch  of  the  rivets  in  one  row;  the 
two  rows  are  '  staggered ',  as  shown  in  Fig,  336,  that  is,  the  rivets 
of  one  row  are  opposite  the  middle  of  the  pitch  of  the  other  row. 

Prob.  33a.  A  boiler  42  inches  in  diameter  carries  a  steam  pressure 
of  1 20  pounds  per  square  inch.  Its  longitudinal  lap  joints  have  a  sin- 
gle row  of  rivets  which  are  spaced  with  ij  inches  pitch.  Compute  the 
tension  P  which  brings  shear  upon  one  rivet. 

Prob.  33b.  The  plates  of  this  boiler  are  }  inches  thick  and  the  rivets 
are  ^J  inches  in  diameter.  Compute  the  unit-stresses  upon  the  plates 
and  the  rivets.  What  is  the  efficiency  of  the  joint,  and  what  steam 
pressure  will  cause  the  boiler  to  rupture  ? 

ART.  34.     DESIGN  OF  RIVETED  JOINTS 

The  design  of  a  riveted  joint  consists  in  giving  such  values 
to  the  plate  thickness  /,  the  diameter  of  the  rivets  d,  and  the 
pitch  of  the  rivets  p,  that  all  parts  may  be  of  the  same  strength, 
and  that  the  working  unit-stresses  may  be  such  that  the  proper 
degree  of  security  is  obtained  (Art.  7).  For  example,  taking  a 
double- riveted  lap  joint  of  structural  steel,  let  it  be  required  to 
determine  /,  d,  and  p  so  that  St=  9000,  Sc  =  12000,  and  Ss  = 
7  500  pounds  per  square  inch  when  the  tension  P  which  comes 
on  one  rivet  is  8  ooo  pounds.  The  third  formula  for  this  case 
(Art.  33)  gives  the  value  of  d,  then  the  second  gives  the  value 


84  CASES  OF  SIMPLE  STRESS  CHAP.  IV 

of  /,  and  finally  the  first  gives  the  value  of  p.  Thus  d2=2P/nS9, 
whence  d  =0.825  inches,  t=P/2dSc=  0.404  inches,  and  p  = 
d+P/tSt  =3.025  inches. 

While  the  above  results  satisfy  the  given  conditions,  it  is 
not  practicable  to  use  the  exact  values,  because  plates  and  rivets 
of  these  dimensions  could  not  be  found  in  the  market  and  would 
have  to  be  specially  made.  The  nearest  approach  to  market 
sizes  would  probably  be  /=f  inches  and  d  =  {-%  inches,  which 
give  _/>  =  3.i8,  or  say  3T\  inches.  Using  these  dimensions,  the 
values  of  Stt  Scy  and  Ss  are  found  to  be  9  ooo,  13  100,  and  7  700 
pounds  per  square  inch,  which  differ  but  little  from  the  speci- 
fied working  unit-stresses. 

Another  method  of  designing  is  to  establish  three  expressions 
for  the  efficiency,  and  then  give  to  /,  d,  and  p  such  values  as 
will  make  the  three  efficiencies  equal  under  the  assigned  unit 
stresses.  Let  a.  denote  the  number  of  rivets  in  the  width  p  which 
transmit  the  tension  P,  and  let  /?  denote  the  number  of  rivet 
sections  in  the  same  space  over  which  the  shear  is  distributed. 
Then  from  the  definition  of  efficiency  in  the  last  article, 

for  tension  of  the  plate,  efficiency  =  (p  —  d)/p 

for  compression  of  the  rivet,    efficiency  =  a  .  dSc/pSt 
for  shear  of  the  rivet,  efficiency  =/?  .  \7td2Ss/ptSt\ 

Equating  now  the  second  of  these  efficiencies  to  the  third,  the 
value  of  d  in  terms  of  /  is  found;  equating  the  first  and  second 
and  eliminating  d,  the  value  of  p  in  terms  of  /  is  obtained; 
accordingly, 

S 


from  which  the  diameter  and  pitch  of  the  rivets  can  be  com- 
puted when  /  is  assumed.  The  efficiency  of  the  joint  now  is 
(p-d)/p  or  aSc/(St+aSe). 

Using  for  steel  plates  and  rivets  the  working  stresses  St  =9  ooo, 
5*c  =  i2ooo,  and  ^=7500  pounds  per  square  inch,  the  above 
formulas  give  for  a  riveted  lap  joint  with  single  riveting,  where 
a  =  i  and  ft  =  i,  the  proportions, 

<f=2.o4/  ^=4.75;  efficiency  =0.57 


ART.  34  DESIGN   OF   RlVETED   JOINTS  85 

so  that  if  the  thickness  of  the  plate  be  given,  and  the  diameter 
and  pitch  of  the  rivets  be  made  according  to  these  rules,  this 
riveted  joint  has  about  57  percent  of  the  strength  of  the  unholed 
plate.  For  a  lap  joint  with  double  riveting,  where  a  =  2  and 
ft  =  2,  the  formulas  become 

d=  2.04!  p=>j.48t  efficiency =0.73 

This  investigation  shows  clearly  the  advantage  of  double  over 
single  riveting,  and  by  adding  a  third  row  the  efficiency  will  be 
raised  to  about  80  percent.  In  both  cases  the  area  t(p-d)  must 
be  sufficient  to  carry  the  tension  P  under  the  unit-stress  St. 

The  application  of  the  above  formulas  to  butt  joints  makes 
the  diameter  of  the  rivet  equal  to  the  thickness  of  the  plate, 
and  makes  the  pitch  much  smaller  than  the  above  values  for 
lap  joints.  These  proportions  are  difficult  to  apply  in  practice 
on  account  of  the  danger  of  injuring  the  metal  in  punching  the 
holes.  For  this  reason  riveted  joints  are  often  made  in  which 
the  strengths  of  the  different  parts  are  not  equal.  Many  other 
reasons,  such  as  cost  of  material  and  facility  of  workmanship, 
influence  also  the  design  of  a  joint.  The  old  rules  which  are 
still  sometimes  used  for  determining  the  pitch  in  butt  joints 
are  expressed  by, 

p=d+^  and  p=d+^ 

the  first  being  for  single  and  the  second  for  double  riveting. 
These  are  deduced  by  making  the  strength  of  the  joint  equal 
in  tension  and  shear,  and  taking  S8=St,  thus  neglecting  entirely 
the  influence  of  the  compression  on  the  rivet. 

The  term  '  bearing  compression  '  is  in  common  use  for  the 
sidewise  compression  brought  upon  a  rivet  by  the  tension  in  the 
plate.  The  assumption  regarding  its  distribution  is  a  rough 
approximate  one  and  rivets  are  more  apt  to  fail  by  shearing  than 
by  bearing  compression.  Hence  it  is  customary  to  allow  a  higher 
working  unit-stress  for  the  bearing  compression  of  a  rivet  than 
for  the  tension  of  a  plate,  notwithstanding  that  rivet  steel  is  gen- 
erally somewhat  lower  in  strength  than  plate  steel  (Art.  25). 
Cooper's  bridge  specifications  give  9  ooo  pounds  per  square 


86  CASES  OF  SIMPLE  STRESS  CHAP.  IV 

inch  as  the  highest  allowable  stress  for  rivets  in  shear  and  15  ooo 
pounds  per  square  inch  for  rivets  in  bearing  compression. 

It  may  be  required  to  arrange  a  joint  so  as  to  secure  either 
strength  or  tightness.  For  a  bridge,  strength  is  mainly  needed; 
for  a  gasholder,  tightness  is  the  principal  requisite;  while  for  a 
boiler  both  these  qualities  are  desirable.  In  general  a  tight 
joint  is  secured  by  using  small  rivets  with  a  small  pitch.  The 
distance  between  the  rows  of  rivets  is  determined  by  practical 
vonsiderations.  The  lap  of  the  plates  should  be  sufficient  so 
chat  the  rivets  may  not  tear  or  shear  the  plate.  If  the  dis- 
tance from  the  center  of  the  hole  to  the  edge  of  the  plate  be  /, 
there  are  two  areas  //  along  which  shearing  tends  to  occur,  and 
2tlS8  must  be  equal  to  or  greater  than  the  tension  P  for  a  single- 
riveted  joint,  or  /  must  be  equal  to  or  greater  than  P/2tS8. 

In  the  preceding  discussion  of  the  stresses  on  the  rivet,  it 
has  been  supposed  that  the  compressive  and  shearing  stresses 
act  independently  of  each  other.  This  assumption  is  the  one 
commonly  made  in  practice,  but  the  investigations  in  Arts.  105 
and  141  show  that  these  stresses  should  really  be  combined  in 
order  to  obtain  the  actual  maximum  stresses  of  compression  and 
shear.  The  usual  method  of  practice  above  explained  in  which 
liberal  factors  of  safety  are  used  is,  however,  generally  regarded 
as  one  giving  ample  security. 

Prob.  34<z.  A  butt  joint  is  like  Fig.  33c,  except  that  it  has  only  one 
cover  plate.  What  should  be  the  thickness  of  this  cover  plate,  and  how 
does  the  joint  differ  from  a  single-riveted  lap  joint  ? 

Prob.  345.  A  lap  joint  with  double  riveting  is  to  be  formed  of  plates 
i  inch  thick  with  rivets  J  inches  diameter.  Find  the  pitch  so  that 
the  strength  of  the  plate  shall  equal  the  shearing  strength  of  the  rivets; 
and  compute  the  efficiency  of  the  structural-steel  joint. 


ART.  35  DEFINITIONS  87 

CHAPTER  V 

GENERAL  THEORY  OF  BEAMS 
ART.  35.     DEFINITIONS 

'  Flexural  Stress  '  occurs  in  a  bar  when  it  is  in  a  horizontal 
position  upon  one  or  more  supports.  The  weight  of  the  bar 
and  its  loads  cause  it  to  bend,  and  induce  in  it  stresses  and  defor- 
mations of  a  complex  nature  which,  as  will  be  seen  later,  may 
be  resolved  into  those  of  tension,  compression,  and  shear.  Such 
a  bar  is  called  a  'Beam  '. 

A  'Simple  Beam  '  is  a  bar  resting  upon  supports  near  its 
ends.  A  '  Cantilever  Beam  '  rests  on  one  support  at  its  middle, 
or  the  portion  of  any  beam  projecting  out  of  a  wall  or  beyond 
a  support  may  be  called  a  cantilever  beam.  A  'Continuous 
3e  im  '  is  a  bar  resting  upon  more  than  two  supports.  In  this 
chapter  the  word  beam,  when  used  without  qualification,  includes 
all  kinds,  whatever  be  the  number  of  the  supports,  or  whether 
the  ends  be  free,  supported,  or  fixed.  Fig.  35a  shows  a  simple 
beam,  Fig.  35Z>  shows  a  cantilever  beam,  and  Fig.  35c  shows  a 
continuous  beam  of  two  equal  spans. 


Fig.  3&z  Fig.  356  Fig.  35c 

The  '  Elastic  Curve  '  is  the  curve  formed  by  a  beam  as  it 
deflects  downward  under  the  action  of  its  own  weight  and  of 
the  loads  upon  it.  Experience  teaches  that  the  amount  of  this 
deflection  and  curvature  is  very  small.  A  beam  is  said  to  be 
'  fixed  '  at  one  end  when  it  is  so  arranged  that  the  tangent  to 
the  elastic  curve  at  that  end  always  remains  horizontal;  this 
may  be  done  in  practice  by  firmly  building  one  end  into  a  wall. 
A  beam  fixed  at  one  end  and  free  at  the  other  end  is  a  canti- 
lever beam. 

The  loads  on  beams  are  either  uniform  or  concentrated.     A 


88  GENERAL  THEORY  OF  BEAMS  CHAP.  V 

'uniform  load'  embraces  the  weight  of  the  beam  itself  and  any 
load  evenly  spread  over  it.  Uniform  loads  are  estimated  by 
their  intensity  per  unit  of  length  of  the  beam,  and  usually  in 
pounds  per  linear  foot.  The  uniform  load  per  linear  unit  is 
designated  by  wt  then  wx  will  represent  the  load  over  any  dis- 
tance MI  if  /  is  the  length  of  the  beam,  the  total  uniform  load 
is  wl,  which  may  be  represented  by  W.  A  '  Concentrated  Load  ' 
is  a  single  applied  weight  and  this  is  designated  by  P;  a  moving 
concentrated  load  is  usually  applied  by  a  rolling  wheel. 

In  this  chapter  the  fundamental  principles  applicable  to  all 
kinds  of  beams  will  be  set  forth.  Unless  otherwise  'stated,  a 
beam  will  be  regarded  as  of  uniform  cross-section  throughout 
its  entire  length,  and  in  computing  its  weight  the  rules  of  Art.  17 
will  be  of  service.  For  example,  the  weight  of  a  wooden  beam 
6x8  inches  in  section  area  and  10  feet  long  is  6x8X3^X1!  = 
133  pounds;  the  weight  of  a  steel  beam  of  the  same  size  is 
6x8X3^X10X1.02  =  1632  pounds. 

Prob.  35a.  Find  the  diameter  of  a  round  steel  bar  which  weighs  6c 
pounds,  its  length  being  5  feet. 

Prob.  356.  A  round  steel  bar  of  2 J  inches  diameter  and  4  feet  length 
weighs  48  pounds.  What  is  the  diameter  of  a  cast-iron  bar  which  has 
the  same  weight  and  length  ? 

ART.  36.     REACTIONS  OF  SUPPORTS 

When  a  beam  is  laid  upon  supports  its  weight  and  the  weight 
of  its  load  are  borne  by  the  supports  which  exert  'reactions  ' 
upward  against  the  beam.  In  fact,  the  beam  is  a  body  in  equilib- 
rium under  the  action  of  a  system  of  forces  which  consists  of 
the  downward  loads  and  the  upward  reactions.  The  loads  are 
usually  given  in  intensity  and  position,  and  it  is  required  to  find 
the  reactions.  This  is  effected  by  the  application  of  the  funda- 
mental conditions  of  static  equilibrium  which,  for  a  system  oi 
vertical  forces  in  one  plane,  are 

Algebraic  sum  of  all  vertical  forces =o 
Algebraic  sum  of  moments  of  all  forces =o 

The  first  of  these  conditions  shows  that  the  sum  of  all  the  loads 


AKT.  36  REACTIONS   OF    SUPPORTS  89 

on  the  beam  is  equal  to  the  sum  of  the  reactions.     Hence  if  there 
/s  but  one  support,  this  condition  gives  at  once  the  reaction. 

For  two  supports  it  is  necessary  to  use  the  second  condition. 
The  axis  of  moments  may  be  taken  at  any  point  in  the  plane, 
but  it  is  more  convenient  to  take  it  at  one  of  the  supports.  For 
example,  consider  a  single  concentrated  load  P  situated  at  4  feet 
from  the  left  end  of  a  simple  beam  whose  span  is  13  feet.  The 
equation  of  moments,  with  the  axis  at  the  left  support,  is 
4P  — 13^2  =  0,  from  which  R2=^P.  Again,  the  equation  of 
moments,  with  the  axis  at  the  right  support,  is  13^1  —  gP=oy 
from  which  Ri=^EP.  As  a  check  it  may  be  observed  that 
Ri+R2  =P. 

For  a  uniform  load  over  a  simple  beam  it  is  evident,  without 
applying  the  conditions  of  equilibrium,  that  each  reaction  is 
one-half  the  load.  For  a  uniform  load  over  the  cantilever  beam 
of  Fig.  356  it  is  plain  that  the  vertical  reaction  at  the  wall  is 
equal  to  the  weight  of  the  load. 


300 
3-- 


1501 
3'—^  ---  4<-- 


fe  4 

Fig.  36a  Fig.  366 

The  reactions  due  to  both  uniform  and  concentrated  loads 
on  a  simple  beam  may  be  obtained  by  adding  together  the  reac- 
tions due  to  the  uniform  load  and  each  concentrated  load,  or 
they  may  be  computed  in  one  operation.  As  an  example  of 
the  latter  method  let  Fig.  366  represent  a  simple  beam  12  feet 
in  length  between  the  supports  and  weighing  35  pounds  per 
linear  foot,  its  total  weight  being  420  pounds.  Let  there  be 
three  concentrated  loads  of  300,  60,  and  150  pounds  placed  at  3, 
5,  and  8  feet  respectively  from  the  left  support.  To  find  the 
right  reaction  R2  the  axis  of  moments  is  taken  at  the  left  sup- 
port, and  the  weight  of  the  beam  regarded  as  concentrated  at 
its  middle;  then  the  equation  of  moments  is, 

=  420X6+300X3  +  60X5+150X8 


90  GENERAL  THEORY  OF  BEAMS  CHAP,  v 

from  which  R2=4io  pounds.  In  like  manner  to  find  RI,  the 
axis  of  moments  is  taken  at  the  right  support,  and 

RiX  12  =  420X6+ 300X9  +  60X7-!-  150X4 

from  which  ^1=520  pounds.  As  a  check  the  sum  of  RI  and 
R2  is  found  to  be  930  pounds,  which  is  the  same  as  the  weight  of 
the  beam  and  the  three  loads. 

When  there  are  more  than  two  supports,  the  problem  of  find- 
ing the  reactions  from  the  principles  of  statics  becomes  inde- 
terminate, since  two  conditions  of  equilibrium  are  only  sufficient 
to  determine  two  unknown  quantities.  By  introducing,  how- 
ever, the  elastic  properties  of  the  material,  the  reactions  of  con- 
tinuous beams  may  be  deduced,  as  will  be  explained  in  a  follow- 
ing chapter.  In  most  cases  of  the  discussion  of  a  beam,  the 
determination  of  the  reactions  is  the  first  step. 

Prob.  36a.  A  simple  beam  weighing  30  pounds  per  linear  foot  is 
18  feet  long,  and  it  carries  two  concentrated  loads  of  350  and  745 
pounds  at  distances  of  yj  and  9  feet  from  the  left  end.  Compute  the 
reactions  due  to  the  total  load. 

Prob.  366.  When  a  single  load  P  is  On  a  simple  beam  of  span  /  at 
a  distance  «/  from  the  left  support,  show  that  the  reactions  are 
Ri  =  P(i-K)  and  R2  =  PK. 

ART.  37.    THE  VERTICAL  SHEAR 

The  failure  of  a  beam  sometimes  occurs  by  shearing  in  a 
vertical  section  as  shown  in  Fig.  37 a  for  a  simple  beam  and  in 
Fig.  37b  for  a  cantilever  beam.  This  shearing  is  produced  by 
two  equal  and  parallel  forces  acting  in  opposite  directions  on 
the  left  and  right  of  the  section.  In  the  second  diagram  there 
acts  on  the  left  of  the  section  a  downward  force  equal  to  the 
sum  of  all  the  loads  on  the  left,  and  on  the  right  of  the  section 
an  equal  force  acts  upward.  In  the  first  diagram  there  acts  on 
the  left  of  the  section  an  upward  force  which  is  equal  to  the 
reaction  minus  the  weight  of  the  beam  between  the  support 
and  the  section.  That  such  forces  actually  exist  will  be  readily 
understood  by  considering  a  numerical  case.  Thus,  for  Fig.  37a, 
let  the  total  weight  of  the  beam  and  loads  between  the  supports 


ART.  37  THE   VERTICAL   SHEAR  91 

be  i  200  pounds,  the  weight  between  the  support  and  the  sec- 
tion mn  be  50  pounds,  and  each  reaction  be  600  pounds;  on 
the  left  of  the  section  mn  the  upward  vertical  force  is  600  —  50  = 
+  550  pounds,  and  on  the  right  of  the  section  the  upward  ver- 
tical force  is  —1150  +  600= —550  pounds;  here  the  sign  + 
indicates  that  the  former  force  acts  upward  and  the  sign  —  indi- 
cates that  the  latter  acts  downward.  In  any  case,  whether  the 
beam  be  simple,  cantilever,  or  continuous,  let  W  indicate  the 
total  weight  upon  the  supports,  W  the  weight  on  the  left  of 
any  section  and  W"  the  weight  on  the  right  of  the  section ,  or 
W'  +  W"=W\  also  let  R'  indicate  the  sum  of  all  the  reactions 
on  the  left  and  R"  the  sum  of  all  the  reactions  on  the  right  of 
the  section,  or  Rf  +  R"  =  W.  Then  the  vertical  force  on  the 
left  of  the  section  is  R'  —  W  and  that  on  the  right  of  the  section 
is  R"  —  W"',  these  two  quantities  are  equal  in  magnitude,  but 
they  have  opposite  signs,  since  their  sum  is  zero;  accordingly 
if  one  acts  upward  the  other  acts  downward  and  they  constitute 
a  shear  (Art.  6). 


„_  __  ,     i;_ 


Fig.  37<z  Fig.  37b 

The  '  Vertical  Shear  '  for  a  section  is  the  name  given  to  the 
algebraic  sum  of  all  the  external  forces  on  the  left  of  the  section. 
Let  upward  forces  be  considered  as  positive  and  downward 
forces  as  negative,  and  let  V  denote  the  vertical  shear  for  a  given 
section;  then, 

V=  Reactions  on  left  of  section  minus  loads  on  left  of  section 
And  the  value  of  V  may  be  positive  or  negative  according  as 
the  reactions  exceed  or  are  less  than  the  loads  on  the  left.  When 
V  is  positive,  the  left-hand  part  of  the  beam  tends  to  slide  upward 
with  respect  to  the  right-hand  part,  as  in  the  section  mn  of  Fig.  370; 
when  V  is  negative,  the  left-hand  part  tends  to  slide  downward 
with  respect  to  the  right-hand  part,  as  in  the  other  section  of 
Fig.  37a.  In  the  cantilever  beam  of  Fig.  376,  there  is  no  reac- 
tion at  the  left,  and  hence  the  vertical  shear  at  every  section  is 
negative. 


92  GENERAL  THEORY  OF  BEAMS  CHAP.  V 

The  vertical  shear  varies  greatly  in  value  at  different  sections 
of  a  beam.  Consider  first  a  simple  beam  of  length  /  and  weigh- 
ing w  per  unit  of  length;  each  reaction  is  then  \wl.  Pass  a 
plane  at  any  distance  x  from  the  left  support;  then  from  the 
definition,  the  vertical  shear  for  that  section  is  V  =  \wl— woe. 
Here  it  is  seen  that  V  has  its  greatest  value  \wl  when  x=o,  that 
V  decreases  as  x  increases,  and  that  V  becomes  o  when  x  =  \l. 
When  x  is  greater  than  J/,  the  value  of  V  is  negative  and  becomes 
—  \wl  when  x=l.  The  equation  V  =  \wl  — wx  is  that  of  a 
straight  line  in  which  x  is  the  abscissa  and  V  the  ordinate,  the 
origin  being  at  the  left  support;  it  may  be  plotted  so  that  the 
ordinate  will  represent  the  vertical  shear  for  the  corresponding 
section  of  the  beam,  as  shown  in  Fig.  37 c. 


Fig.  37c  Fig.  37 d 

Consider  again  a  simple  beam,  as  in  Fig.  37 d,  having  a  span 
of  12  feet  with  three  loads  of  240,  90,  and  120  pounds,  situated 
3,  4,  and  8  feet  respectively  from  the  left  support.  By  Art.  36 
the  left  reaction  is  found  to  be  280  and  the  right  reaction  170 
pounds.  Then,  for  any  section  between  the  left  support  and 
the  first  load  the  vertical  shear  due  to  the  given  loads  is  V  ==  +280 
pounds,  for  a  section  between  the  first  and  second  loads  it  is 
7  =  280  —  240=4-40  pounds,  for  a  section  between  the  second 
and  third  loads  it  is  7  =  280-240-90= -50  pounds,  and  for 
a  section  between  the  third  load  and  the  right  support  it  is  V  = 
280  — 240  — 90  — 120= —170  pounds,  which  has  the  same  numer- 
ical value  as  the  right  reaction.  By  laying  off  ordinates  upon  a 
horizontal  line  a  graphical  representation  of  the  vertical  shears 
due  to  the  three  concentrated  loads  is  obtained. 

For  any  section  of  a  simple  beam  distant  x  from  the  left 
support,  let  RI  denote  the  left  reaction,  w  the  weight  of  the  uni- 
form load  per  linear  unit,  and  IP\  the  sum  of  all  the  concen- 
trated loads  between  the  section  and  the  support.  Then  the 


ART.  38  THE    BENDING   MOMENT  93 

definition  gives  V  =  Ri—wx  —  2Pi  as  a  general  expression  for  the 
vertical  shear.  Thus  if  the  beam  of  the  last  paragraph  weighs 
20  pounds  per  linear  foot,  the  vertical  shear  at  the  left  support 
is  400  pounds,  while  at  two  feet  from  the  left  support  it  is  400  — 
40  =  360  pounds.  The  student  should  compute  vertical  shear? 
for  other  sections  and  draw  the  shear  diagram. 

The  vertical  shear  for  any  section  of  a  beam  is  a  measure  oi 
the  tendency  to  shearing  along  that  section.  The  above  examples 
show  that  this  is  greatest  near  the  supports.  It  is  rare  that 
beams  actually  fail  in  this  manner,  but  it  is  often  necessary  to 
investigate  the  tendency  to  such  failure. 

Prob.  37.  The  cantilever  beam  in  Fig.  38a  is  10  feet  long  and 
weighs  23  pounds  per  linear  foot,  while  the  load  P  is  60  pounds  and 
is  placed  at  4  feet  from  the  wall.  Compute  the  vertical  shears  at 
several  sections  throughout  the  beam,  and  draw  a  diagram  to  show 
their  distribution. 

ART.  38.     THE  BENDING  MOMENT 

The  usual  method  of  failure  of  beams  is  by  cross-breaking  or 
transverse  rupture.  This  is  caused  by  the  external  forces  pro- 
ducing rotation  around  some  point  in  the  section  of  failure. 
Thus,  in  Fig.  386  let  /  be  the  length  of  the  cantilever  beam  and 
let  m  be  the  distance  between  P  and  the  wall.  Then  the  ten- 
dency of  P  to  cause  rotation  around  a  point  in  the  section  at 
the  wall  is  measured  by  its  moment  —  Pm ;  if,  however,  the  load 
is  at  the  end,  its  tendency  to  produce  rotation  around  the  same 
point  is  measured  by  the  moment  —  PI.  If  w  is  the  weight  of 
this  beam  per  linear  unit,  the  uniform  load  wl  produces  the 
same  tendency  to  rotation  as  if  it  were  concentrated  at  its  center 
of  gravity  (Elements  of  Mechanics,  Art.  13);  hence  with  respect 
to  a  section  at  the  wall  the  moment  of  the  uniform  load  wl 
is  -wlXty. 

Moments  are  taken  as  positive  when  they  tend  to  cause  rota- 
tion in  the  same  direction  as  the  hands  of  a  clock,  and  negative 
when  they  tend  to  cause  rotation  in  the  opposite  direction.  When 
the  force  is  in  pounds  and  the  lever  arm  is  in  feet,  the  moment 


94 


GENERAL  THEORY  OF  BEAMS 


CHAP.  V 


is  in  pounds-feet.  For  instance,  let  the  beam  of  Fig.  38b  be 
10  feet  long  and  weigh  23  pounds  per  linear  foot,  and  let  the 
concentrated  load  P  be  60  pounds  and  be  placed  at  6  feet  from 
the  end.  Then,  the  moment  under  the  load  is  -138X3=  -414 
pound-feet,  and  that  at  the  wall  is  —230X5— 60X4= —1390 
pound-feet. 


Fig.  38a  Fig.  386 

The  algebraic  sum  of  the  moments  of  all  the  external  forces 
on  the  left  of  any  section  in  a  beam  is  called  the  'Bending 
Moment '  for  that  section.  These  external  forces  consist  of 
upward  reactions  and  downward  loads,  and  hence  the  moment 
of  any  reaction  is  positive  and  that  of  any  load  is  negative.  Let 
the  bending  moment  be  designated  by  M;  then  for  any  section, 

M  =  sum  of  moments  of  reactions  minus  sum  of  moments  of  loads 

and  the  bending  moment  may  be  positive  or  negative  according 
as  the  first  or  second  term  is  the  greater.  For  the  cantilever 
beam,  illustrated  above,  there  is  no  reaction  at  the  left  end,  and 
hence  all  the  bending  moments  are  negative. 

For  a  simple  beam  of  length  /,  uniformly  loaded  with 
iv  per  linear  unit,  each  reaction  is  \wl.  For  any  section  dis- 
tant x  from  the  left  support,  the  bending  moment  is  M  = 
\wl.x-wx.\x,  x  being  the  lever  arm  of  the  reaction  \wl> 
and  \x  the  lever  arm  of  the  load  wx.  Here  M=o  when  x  =  o 
and  also  when  x=l,  and  M  has  its  greatest  value  Jw/2  when 
x  =  %l.  The  equation  is  that  of  a  parabola  whose  graphical 
representation  is  as  given  in  Fig.  38d,  each  ordinate  show- 
ing the  value  of  M  for  the  corresponding  value  of  the 
abscissa  x.  Consider  next  a  simple  beam  loaded  with  only 
the  three  weights  PI,  P%,  P$.  Here  M=R\x  for  any  sec- 


ART.  38 


THE  BENDING  MOMENT 


95 


tion  between  the  left  support  and  the  first  load,  and  M  = 
RiX-Pi(x  —  pi)  for  any  section  between  the  first  and  second 
loads.  Each  of  these  expressions  is  the  equation  of  a  straight  line, 
x  being  the  abscissa  and  M  the  ordinate,  and  the  graphical  repre- 
sentation of  bending  moments  is  as  shown  in  Fig.  3Sc.  It  is  seen 
that  for  a  simple  beam  all  the  bending  moments  are  positive. 

For  any  given  case,  the  bending  moment  at  any  section  may 
be  found  by  using  the  above  definition.  The  external  forces 
on  the  left  of  the  section  are  taken  merely  for  convenience,  for 
•hose  upon  the  right  of  the  section  produce  the  same  bending 
moment  with  reference  to  the  section.  The  bending  moment 
in  all  cases  is  a  measure  of  the  tendency  of  the  external  forces 
on  either  side  of  the  section  to  turn  around  a  point  in  that  section. 
If  this  bending  moment  is  sufficiently  large,  the  internal  stresses 
in  the  beam  become  so  great  that  rupture  occurs. 


Fig.  38c  Fig.  38d 

The  bending  moment  at  any  section  may  also  be  obtained 
by  using  the  forces  on  the  right  of  that  section.  Thus,  for  the 
beam  in  Fig.  38c,  let  R2  be  the  right  reaction  and  ps  the 
distance  of  the  load  PS  from  the  right  support;  then  for  a 
section  under  the  load  PS  the  bending  moment  is  M  =  R2ps  and 
this  is  equal  to  that  found  by  using  the  forces  on  the  left  of 
the  section.  When  forces  on  the  right  of  a  section  are  used,  a 
moment  is  to  be  taken  as  positive  when  it  tends  to  cause 
rotation  in  the  opposite  direction  to  that  of  the  hands  of  a 
clock,  for  this  direction  would  be  clockwise  if  the  beam  be 
viewed  from  the  other  side. 

Prob.  38.  Draw  a  beam  resting  on  three  supports  and  having  two 
spans  each  13  feet  long.  Let  a  load  of  160  pounds  be  placed  at  the 
middle  of  each  span  producing  a  reaction  of  220  pounds  at  the  middle 
support  and  50  pounds  at  each  end  support.  Compute  the  vertical 


96  GENERAL  THEORY  OF  BEAMS  CHAP,  v 

shears  and  bending  moments  for  several  sections,  and  draw  diagrams 
to  show  their  variation  throughout  the  beam. 

ART.  39.    INTERNAL  STRESSES  AND  EXTERNAL  FORCES 

The  external  loads  and  reactions  on  a  beam  maintain  their 
equilibrium  by  means  of  internal  stresses  which  are  generated 
in  it.  It  is  required  to  determine  the  relations  between  the 
external  forces  and  the  internal  stresses;  or,  since  the  effect  of 
the  external  forces  upon  any  section  is  represented  by  the  ver- 
tical shear  (Art.  37)  and  by  the  bending  moment  (Art.  38),  the 
problem  is  to  find  the  relation  between  these  quantities  and  the 
internal  stresses  in  that  section. 

i        I     mi         j  j         I     m  j  ^X 


V?      Y 


Fig.  39<z  Fig.  396 

Consider  a  beam  of  any  kind  which  is  loaded  in  any  manner. 
Imagine  a  vertical  plane  cutting  the  beam  at  any  section  mn, 
as  in  Fig.  39a.  In  that  section  there  are  acting  unknown  stresses 
of  various  intensities  and  directions.  Let  the  beam  be  imagined 
to  be  separated  into  two  parts  by  the  cutting  plane  and  let  forces 
X,  Y,  Z,  etc.,  equivalent  to  the  internal  stresses,  be  applied  to 
the  section  as  shown  in  Fig.  396.  Then  the  equilibrium  of  each 
part  of  the  beam  will  be  undisturbed,  for  each  part  will  be  acted 
upon  by  a  system  of  forces  in  equilibrium.  Hence  the  following 
fundamental  principle  is  established. 

The  internal  stresses  in  any  cross-section  of  a  beam  hold  in 
equilibrium  the  external  forces  on  each  side  of  that  section. 

This  is  the  most  important  principle  in  the  theory  of  flexure. 
It  applies  to  all  beams,  whether  the  cross-section  be  uniform  or 
variable  and  whatever  be  the  number  of  the  spans  or  the  nature 
of  the  loading. 

In  the  above  figure  the  internal  stresses  X,  Y,  Z,  etc.,  hold 
in  equilibrium  the  loads  and  reactions  on  the  left  of  the  section, 
and  also  those  on  the  right.  Considering  one  part  only,  a  sys- 
tem of  forces  in  equilibrium  is  seen,  to  which  the  three  necessary 


ART.  39      INTERNAL  STRESSES  AND  EXTERNAL  FORCES  97 

and  sufficient  conditions  of  statics  for  forces  in  one  plane  apply 
(Elements  of  Mechanics,  Art.  5),  namely, 

Algebraic  sum  of  all  horizontal  components =o 
Algebraic  sum  of  all  vertical  components =o 
Algebraic  sum  of  moments  of  all  forces =o 

From  these  conditions  can  be  deduced  three  laws  concerning 
the  unknown  stresses  in  any  section.  Whatever  be  the  inten- 
sity and  direction  of  these  ,  v  II 

stresses,  let  each  be  resolved    * , —  — : 1 — L 

into  its  horizontal  and  verti- 
cal components.  The  horizon-  ^ 
tal  components  will  be  ap-  j 
plied  at  different  points  in  the  Fig.  39c 
cross-section,  some  acting  in  one  direction  and  some  in  the 
other,  or  in  other  words,  some  of  the  horizontal  stresses  are  ten- 
sile and  some  compressive;  by  the  first  condition  the  algebraic 
sum  of  these  is  zero.  The  vertical  components  will  add  together 
and  form  a  resultant  vertical  force  V  which,  by  the  second  con- 
dition, equals  the  algebraic  sum  of  the  external  forces  on  the 
left  of  the  section.  Since  this  internal  force  V  acts  in  contrary 
directions  upon  the  two  parts  into  which  the  beam  is  supposed 
to  be  separated,  it  is  of  the  nature  of  a  shear.  Hence  for  any 
section  of  any  beam  the  following  laws  concerning  the  internal 
stresses  may  be  stated. 

1.  The  algebraic  sum  of  the  horizontal  stresses  is  zero;  or 
the  sum  of  the  horizontal  tensile  stresses  is  equal  to  the  sum  of 
the  horizontal  compressive  stresses. 

2.  The  algebraic  sum  of  the  vertical  stresses  forms  a  re- 
sultant shear  which  is  equal  to  the  algebraic  sum  of  the  external 
vertical  forces  on  either  side  of  the  section. 

3.  The  algebraic  sum  of  the  moments  of  the  internal  stresses 
is  equal  to  the  algebraic  sum  of  the  moments  of  the  external 
forces  on  either  side  of  the  section. 

These  three  theoretical  laws  are  the  foundation  of  the  theory 
of  the  flexure  of  beams;  they  may  be  expressed  in  simpler  form 
by  the  help  of  the  following  definitions. 


98  GENERAL  THEORY  OF  BEAMS  CHAP.  V 

*  Resisting  shear  '  is  the  name  given  to  the  algebraic  sum  of 
the  internal  vertical  stresses  in  any  section,  and  '  vertical  shear  ' 
is  the  name  for  the  algebraic  sum  of  the  external  vertical  forces 
on  the  left  of  the  section.  'Resisting  moment '  is  the  name 
given  to  the  algebraic  sum  of  the  moments  of  the  internal  hori- 
zontal stresses  with  reference  to  a  point  in  the  section,  and 
'bending  moment  '  is  the  name  for  the  algebraic  sum  of  the 
moments  of  the  external  forces  on  either  side  of  the  section  with 
reference  to  the  same  point.  Then  the  three  laws  may  be  thus 
expressed  for  any  section  of  any  beam : 

Sum  of  tensile  stresses  =  Sum  of  compressive  stresses 
Resisting  shear  =  Vertical  shear 

Resisting  moment         =  Bending  moment 

The  second  and  third  of  these  equations  furnish  the  funda- 
mental laws  for  investigating  beams.  They  state  the  relations 
between  the  internal  stresses  in  any  section  and  the  external 
forces  on  either  side  of  that  section.  For  the  sake  of  uniformity 
the  external  forces  on  the  left-hand  side  of  the  section  will  gen- 
erally be  used,  as  was  done  in  Arts.  37  and  38. 

Prob.  39.  A  beam  6  feet  long  is  sustained  at  one  end  by  a  force  of 
280  pounds  acting  at  an  angle  of  60  degrees  with  the  vertical,  and  at 
the  other  end  by  a  vertical  force  Y  and  a  horizontal  force  X.  Find 
the  values  of  X  and  Y,  and  the  weight  of  the  beam. 

ART.  40.     NEUTRAL  SURFACE  AND  Axis 

From  the  three  necessary  conditions  of  static  equilibrium,  as 
stated  in  Art.  39,  three  important  theoretical  laws  regarding 
internal  stresses  were  deduced.  These  alone,  however,  are  not 
sufficient  for  the  full  investigation  of  the  subject,  but  recourse 
must  be  had  to  experience  and  experiment.  Experience  teaches 
that  when  a  beam  deflects,  one  side  becomes  concave  and  the 
other  convex,  and  it  is  reasonable  to  suppose  that  the  horizontal 
tensile  stresses  are  on  the  convex  side  and  the  compressive  stresses 
on  the  concave  side.  By  experiments  on  beams  this  is  confirmed, 
and  it  is  also  found  that  any  two  parallel  vertical  straight  lines 
drawn  on  the  beam  before  flexure  remain  straight  after  flexure, 


ART.  40  NEUTRAL  SURFACE  AND  Axis  99 

but  are  nearer  together  than  before  on  the  compressive  side  and 
farther  apart  on  the  tensile  side.  Accordingly  the  following 
experimental  laws  may  be  stated : 

4.  The  horizontal  fibers  on  the  convex  side  are  elongated  and 
those  on  the  concave  side  are  shortened,  while  near  the  cen- 
ter there  is  a  *  neutral  surface'  which  is  unchanged  in  length. 

5.  The  elongation  or  shortening  of  any  fiber  is  directly  pro- 
portional to  its  distance  from  the  neutral  surface. 

Now  when  the  elastic  limit  of  the  material  is  not  exceeded,  the 
stresses  in  the  fibers  are  proportional  to  their  changes  in  length 
(Art.  2);  therefore, 

6.  The  horizontal  stresses  are  directly  proportional  to  their 
distances  from  the  neutral  surface,  provided  all  unit-stresses 
are  less  than  the  elastic  limit  of  the  material. 

From  these  laws  there  will  now  be  deduced  the  following  important 
theorem  regarding  the  position  of  the  neutral  surface: 

The  neutral  surface  passes  through  the  centers  of  gravity 
of  the  cross-sections. 

To  prove  this  let  da  be  the  area  of  any  elementary  fiber  and  z 
its  distance  from  the  neutral  surface.  Let  S  be  the  unit-stress 
on  the  horizontal  fiber  most  remote  from  the  neutral  surface  at 
the  distance  c.  Then  by  the  sixth  law, 

S/c= unit-stress  at  the  distance  unity, 
S  .  z/c= unit-stress  at  the  distance  z. 

The  total  horizontal  stress  on  the  fiber  at  the  distance  z  now 
is  da  .  Sz/c,  and  hence  for  the  entire  cross-section, 

2 da  .  Sz/c=  (S/c)2da  .  z= algebraic  sum  of  all  horizontal  stresses. 

But  by  the  first  law  of  Art.  39  this  algebraic  sum  is  zero,  and 
hence  the  quantity  Ida  .  z  must  be  zero.  This,  however,  is  the 
condition  which  makes  the  line  of  reference  pass  through  the 
center  of  gravity,  as  is  plain  from  \~  ~n  f 


Neutral  Sitrfnre 


the  definition  of  the  term  '  center 
of  gravity '  (Elements  of  Me- 
chanics, Art.  13).  Therefore  the 
neutral  surface  of  a  beam  passes  Fig.  40 

through  the    centers    of   gravity    of   the    cross-sections. 


100  GENERAL  THEORY  OF  BEAMS  CHAP.  \ 

The  'neutral  axis'  of  a  cross-section  is  the  line  in  which  the 
neutral  surface  intersects  the  plane  of  the  cross-section.  On 
the  left  of  Fig.  40  is  shown  the  neutral  axis  of  a  cross-section 
and  on  the  right  a  trace  of  the  neutral  surface. 

When  a  beam  is  loaded  so  heavily  that  the  horizontal  unit- 
stress  in  any  fiber  exceeds  the  elastic  limit  of  the  material,  the 
neutral  surface  no  longer  passes  through  the  centers  of  gravity 
of  the  cross-sections  since  the  fifth  law  is  no  longer  correct. 
The  common  theory  of  flexure,  developed  in  this  and  the  follow- 
ing chapters,  applies  therefore  only  to  cases  in  which  the  unit- 
stresses  are  less  than  the  elastic  limit. 

Let  S  be  the  unit-stress  on  the  horizontal  fiber  at  the  most 
remote  distance  c  from  the  neutral  surface  and  S\  be  the  unit- 
stress  on  a  fiber  at  the  distance  c\.  Then  the  sixth  law  gives, 

Si/S=ci/c  or  Si/ci  =  S/c  (40) 

For  example  in  the  beam  of  Fig.  40  let  the  depth  be  9  inches 
and  the  neutral  surface  be  3^  inches  from  the  base;  let  the  hori- 
zontal unit-stress  at  the  upper  surface  be  4  200  pounds  per  square 
inch;  then  the  horizontal  unit-stress  at  the  lower  surface  is 
Si  =S  .  Ci/c  =4  200X3.5/5.5  =2  670  pounds  per  square  inch. 
Si  is  tension  when  S  is  compression,  and  Si  is  compression  when 
S  is  tension,  for  the  stresses  on  opposite  sides  of  the  neutral  sur- 
face must  be  of  different  kinds  since  their  algebraic  sum  is  zero. 

The  sum  of  all  the  horizontal  stresses  above  or  below  the 
neutral  axis  is  (S/c)Ida  .  z,  in  which  Ida  .  z  is  equal  to  the 
moment  of  the  area  of  the  cross-section  above  or  below  the  neu- 
tral axis  with  respect  to  that  axis.  Thus,  for  a  beam  4  inches 
wide  and  6  inches  deep  the  section  area  above  the  neutral  axis 
is  4X3  =  12  square  inches,  and  its  center  of  gravity  is  ij  inches 
from  that  axis,  so  that  the  value  of  Ida  .  z  is  12 X i J  =  18  inches3; 
if  S  for  this  case  is  600  pounds  per  square  inch,  the  sum  of  all 
the  stresses  above  the  neutral  axis  is  (600/3)  X  18=3  600  pounds. 
This  result  may  also  be  obtained  in  another  way:  the  section 
area  above  the  neutral  axis  is  12  square  inches  and  the  mean 
unit-stress  upon  it  is  300  pounds  per  square  inch;  hence  the 
total  stress  is  12X300  =  3  600  pounds. 


ART  41  SHEAR  AND  FLEXURE  FORMULAS  101 

Prob.  40.  Let  Fig.  40  represent  the  section  of  a  cast-iron  beam  in 
which  c  is  5  inches,  the  thickness  of  the  web  ij  inches,  the  width  of 
the  upper  flange  3  inches,  and  its  depth  2  inches.  If  the  horizontal 
unit-stress  on  the  upper  fiber  is  6  700  pounds  per  square  inch,  com- 
pute the  total  horizontal  stress  above  the  neutral  axis. 

ART.  41.     SHEAR  AND  FLEXURE  FORMULAS 

Consider  again  any  beam  loaded  in  any  manner  and  cut  at 
any  section  by  a  vertical  plane.  The  internal  stresses  in  that 
section  hold  in  equilibrium  the  external  reactions  and  loads  on 
the  left  of  the  section,  and  as  shown  in  Art.  39,  the  following 
fundamental  equations  apply  to  that  section: 

Resisting  shear      =  Vertical  shear 

Resisting  moment=  Bending  moment 

The  principles  established  in  the  preceding  pages  can  now  be 
applied  to  the  algebraic  expression  of  these  four  quantities. 

The  resisting  shear  is  the  algebraic  sum  of  all  the  vertical 
components  of  the  internal  stresses  at  any  section  of  the  beam. 
If  a  is  the  area  of  that  section  and  S8  the  shearing  unit-stress, 
regarded  as  uniform  over  the  section  area,  then  from  Art.  6, 

Resisting  shear  =aSs 

The  vertical  shear  for  the  same  section  of  the  beam  being  V 
(Art.  37),  the  first  of  the  above  equations  becomes, 

Ssa=V          or  Ss=V/a 

which  is  the  fundamental  formula  for  the  discussion  of  shearing 
stresses  in  beams;  this  will  hereafter  generally  be  called  the 
'shear  formula ',  and  it  assumes  that  the  shear  is  uniformly  dis- 
tributed over  the  section  area. 

The  resisting  moment  is  the  algebraic  sum  of  the  moments 
of  the  internal  horizontal  stresses  in  any  section  with  reference 
to  a  point  in  that  section.  To  find  an  expression  for  its  value 
let  5  be  the  horizontal  unit-stress,  tensile  or  compressive  as  the 
case  may  be,  upon  the  fiber  most  remote  from  the  neutral  axis, 
and  let  c  be  the  shortest  distance  from  that  fiber  to  said  axis. 
Also  let  z  be  the  distance  from  the  neutral  axis  to  any  fiber  having 
the  elementary  area  da.  Then  by  Art.  40, 


102         GENERAL  THEORY  OF  BEAMS       CHAP  V 

S/c= unit-stress  at  distance  unity 
S  .  z/c= unit-stress  at  distance  z 
da  .  Sz/c=  stress  on  fiber  of  area  da 

The  moment  of  this  fiber  stress  with  respect  to  the  neutral  axis 
of  the  cross-section  is  da  .  Sz2/c,  and  the  algebraic  sum  of  all 
the  elementary  moments  is  the  resisting  moment,  or 

(S/c}2da  .  z2= resisting  moment  of  horizontal  stresses 
But  the  quantity  Ida  .  z2,  being  the  sum  of  the  products  formed  by 
multiplying  each  elementary  area  by  the  square  of  its  distance 
from  the  neutral  axis,  is  the  '  moment  of  inertia '  of  the  area  of  the 
cross-section  with  reference  to  that  axis.  Let  this  moment  of 
inertia  be  represented  by  / ;  then, 

Resisting  moment  =  (S/c)I=S  .  I/c 

The  bending  moment  for  the  same  section  of  the  beam  being  M 
(Art.  38),  the  second  of  the  above  equations  becomes, 

S.I/c=M  or  S=M.c/I  (41) 

which  is  the  fundamental  formula  for  the  discussion  of  the  hori- 
zontal tensile  and  compressive  stresses  in  beams;  this  will  here- 
after generally  be  called  the  'flexure  formula'. 

Experience  and  experiment  teach  that  simple  beams  of  uni- 
form section  break  near  the  middle  by  the  tearing  or  crushing 
of  the  fibers  and  rarely  at  the  supports  by  shearing.  Hence  it 
is  the  flexure  formula  that  is  mostly  employed  in  the  practical 
investigation  of  beams,  although  for  short  beams  the  shear  formula 
must  also  be  used.  When  the  beam  is  of  varying  cross-section^ 
as  is  the  case  in  plate  girders,  the  two  formulas  are  of  equal 
importance, 

When  a  beam  of  given  size  and  shape  is  to  be  discussed, 
its  dimensions  furnish  the  value  of  the  section  area  a,  the  moment 
of  inertia  /,  and  the  distance  c  from  the  neutral  axis  of  the  sec- 
tion to  the  side  of  the  beam  furthest  from  that  axis.  The 
quantity  I/c  hence  depends  only  on  the  dimensions  of  the  cross- 
section  and  not  ai  ill  upon  the  loads  or  the  material  of  the  beam ; 
for  this  reason  it  ;s  called  the  'section  factor.'  There  is  a 
certain  analogy,  then,  between  the  two  fundamental  formulas 
for  beams;  from  the  first  the  value  of  V/S8  must  equal  the 


ART.  42  CENTERS    OF    GRAVITY  103 

section  area  0,  from  the  second  the  value  of  M/S  must  equal 
the  section  factor  I/c. 

To  determine  the  shearing  unit-stress  in  a  given  section  of 
a  beam  due  to  given  loads,  the  vertical  shear  V  is  computed 
by  Art.  37  and  the  section  area  a  by  the  rules  of  geometry;  then 
the  shear  formula  gives  S8  =  V/a.  To  determine  the  greatest 
unit-stress  of  tension  or  compression  in  the  given  section,  the 
bending  moment  M  is  computed  by  Art.  38,  and  the  quantities 
c  and  /  by  the  methods  of  Arts.  42  and  43;  then  the  flexure 
formula  gives  S  =  M  .  c/I.  Many  applications  of  these  formulas 
to  various  kinds  of  beams  will  be  found  in  the  following  chapters. 

Prob.  41.  A  simple  beam  has  a  section  area  of  20  square  inches 
and  a  span  of  15  feet.  Over  it  roll  two  locomotive  wheels  6  feet  apart 
and  each  bringing  12  ooo  pounds  upon  the  beam.  Find  the  position 
of  these  Wheels  so  as  to  give  the  greatest  vertical  shear  at  a  section  close 
to  one  of  the  supports.  Compute  the  mean  shearing  unit-stress  in 
that  section. 

ART.  42.     CENTERS  OF  GRAVITY 

In  the  flexure  formula  S  .  I/c=M,  the  quantity  c  is  the  short- 
est distance  from  the  remotest  part  of  the  cross-section  to  a  hori- 
zontal axis  passing  through  the  center  of  gravity  of  that  section. 
Whenever  a  cross-section  is  symmetric  with  respect  to  this  axis, 
as  in  all  the  cases  of  Fig.  420,  the  center  of  gravity  is  evidently 
at  the  middle  of  the  depth;  thus,  if  d  is  the  depth  of  the  section 
the  value  of  c  is  d. 


Fig.  42a 

For  unsymmetric  sections  the  value  of  c  is  to  be  computed 
from  the  definition  of  the  center  of  gravity,  using  the  principle 
of  moments  (Elements  of  Mechanics,  Art.  15).  For  example, 
take  the  T  section  shown  in  Fig.  42b,  the  depth  being  d}  the 
width  of  the  top  flange  6,  the  thickness  of  that  flange  /i,  and 
the  thickness  of  the  web  t\  the  area  of  the  section  is  a  =td+ti(b  —  /). 


104 


GENERAL  THEORY  OF  BEAMS 


CHAP.  V 


Taking  an  axis  of  moments  at  the  foot  of  the  web,  the  equation 
of  moments  is 


from  which  c  is  known.  For  example,  let  b—  5,  /  =  J,  t\  =\  and 
d=4  inches;  then  0  =  4.25  square  inches,  and  £  =  2.93  inches. 
The  same  formula  holds  for  the  section  in  Fig.  42c.  For  the 
sections  in  Figs.  42d  and  420,  let  d  be  the  depth,  bi  and  /i  the 
width  and  thickness  of  the  larger  flange,  b2  and  /2  the  -width 
and  thickness  of  the  smaller  flange,  and  /  the  thickness  of  the 
web;  then  the  area  of  the  section  is  a  =  td+ti(bi—  /)  +  /2(&2~0> 
and  the  equation  of  moments  with  respect  to  the  bottom  of  the 
lower  flange  is, 


from  which  c  is  computed.  ,  Actual  sections  of  these  forms  have 
the  corners  more  or  less  rounded,  but  values  of  c  computed  in 
this  manner  are  usually  sufficiently  precise  for  all  practical  uses. 


<---&-,--* 


Fig.  426 


Fig.  42c 


Fig.  42d 


Fig. 


The  student  should  be  prepared  to  readily  apply  the  prin- 
ciple of  moments  to  the  deduction  of  the  numerical  value  of  c 
for  any  given  cross-section.  In  nearly  all  cases  the  given  area 
may  be  divided  into  rectangles,  triangles,  and  circular  sectors, 
the  centers  of  gravity  of  which  are  known,  so  that  the  equation 
for  finding  c  is  readily  written  from  the  definition  of  the  center 
of  gravity.  Triangular  beams  are  never  used  in  practice,  but 
it  is  well  to  note  that  the  center  of  gravity  of  a  triangle  lies  upon 
a  line  parallel  to  its  base  and  at  a  distance  from  that  base  equal 
to  one-third  of  the  altitude. 

Centers  of  gravity  for  sections  like  those  of  railroad  rails 
are  determined  by  dividing  the  area  into  strips  which  are  so 
small  that  each  may  be  regarded  as  a  trapezoid.  The  areas  of 


ART.  43  MOMENTS   OF   INERTIA  105 

these  elementary  trapezoids  being  computed,  their  centers  of 
gravity  may  be  taken  at  the  middle  of  their  widths  with  sufficient 
precision,  and  then  the  value  of  c  is  determined  by  the  same 
method  as  that  explained  above.  There  is  also  a  graphical 
method  for  locating  the  center  of  gravity  of  sections  which  is 
sometimes  advantageous,  and  this  is  set  forth  in  Roofs  and 
Bridges,  Part  II,  Art.  12. 

Prob.  42a.  If  the  side  of  a  square  is  d,  show  that  c=dVr%  when  the 
square  is  placed  so  that  one  diagonal  is  vertical. 

Prob.  426.  For  a  trapezoidal  section  of  depth  d,  let  the  shorter  base 
be  b'  and  the  longer  base  be  b.  Find  the  distance  of  the  center  of 
gravity  from  the  shorter  base. 

Prob.  42c.  A  deck  beam  has  a  section  like  Fig.  426,  except  that 
the  lower  flange  is  approximately  circular.  Find  the  value  of  c  when 
the  total  depth  is  10  inches,  the  width  of  the  upper  flange  5  inches, 
the  thickness  of  that  flange  J  inch,  the  thickness  of  the  web  f  inch,  and 
the  diameter  of  the  circular  bulb  i  j  inches. 

ART.  43.    MOMENTS  OF  INERTIA 

In  the  flexure  formula  S  .I/c=M,  the  letter  /  denotes  the 
moment  of  inertia  of  the  cross-section  of  the  beam  with  refer- 
ence to  a  horizontal  axis  passing  through  the  center  of  gravity 
of  that  section.  Let  da  be  any  elementary  area  of  the  section 
and  z  its  distance  from  this  axis;  then  I  =  Zdz  .  z2  is  the  equa- 
tion from  which  the  value  of  /  is  ascertained,  and  this  in  general 
is  most  advantageously  done  by  the  methods  of  the  calculus. 
Strictly  speaking  an  area  has  no  weight  or  inertia  and  therefore 
no  moment  of  inertia,  but  it  is  customary  to  give  the  name 
'moment  of  inertia  '  to  the  quantity  Ida  .  z2,  which  is  of  very 
frequent  occurrence  in  all  branches  of  applied  mechanics. 

For  the  rectangle  of  breadth  b  and  depth  d  in  Fig.  43a,  the 
elementary  area  da  may  be  taken  as  a  strip  of  length  b  and  width 
dz,  or  da  =  b  .  dz.  Then  Ida  .  z2  equals  fbz2dz  and  the  value 
of  z  in  this  integral  is  to  extend  between  the  limits  +  \d  and  -  \d. 
This  gives  I  =  ^bd3,  which  is  the  moment  of  inertia  of  the  rect- 


106  GENERAL  THEORY  OF  BEAMS  CHAP.  V 

angle  with  respect  to  an  axis  through  its  center  of  gravity  and 
parallel  to  the  base. 

For  the  triangle  with  base  b  and  altitude  d  in  Fig.  43b,  the 
elementary  strip  has  the  width  dz  and  the  variable  length  x. 
Since  the  center  of  gravity  is  at  a  distance  of  %d  from  the  base, 
the  value  of  x  is,  from  similar  triangles,  x  =  ($d—z)b/d,  and 
the  expression  Ida  .  z2  becomes  b/d  .  J(%d-z)z2dz.  This  being 
taken  between  the  limits  +%d  and  —$d,  gives  the  result  /  = 
-s^bd3,  which  is  the  moment  of  inertia  of  the  triangle  with  respect 
to  an  axis  through  its  center  of  gravity  and  parallel  to  the  base  b. 

For  the  circle  of  diameter  d  in  Fig.  43c,  a  similar  method 
may  be  used.  The  length  of  the  elementary  strip  being  2X9 
the  relation  between  x  and  z  is  given  by  x2  +  z2=(^d)2.  The 
general  expression  2 da  .  z2  becomes  J  2xz2dz,  and,  replacing  x 
Sy  its  value  in  terms  of  z,  and  integrating  between  the  limits 
+  Jdand  —  %d,  gives  the  result  /=  fand*,  which  is  the  moment  of 
inertia  of  the  area  of  a  circle  with  respect  to  an  axis  through 
its  center. 


Fig.  43a  Fig.  436  Fig.  43c  Fig.  43d 

For  the  circular  annulus  in  Fig.  43 d,  let  d  be  the  outer  and 
di  the  inner  diameter,  then  it  is  plain  from  the  definition  that 
the  moment  of  inertia  of  the  annulus  is  that  of  the  outer  circle 
diminished  by  that  of  the  inner  circle  or  I  =  -fax(d4- d^). 
Similarly  for  the  hollow  rectangle  in  Fig.  43e,  let  b  and  61  be 
the  widths  and  d  and  d\  the  depths,  the  thickness  of  both  top 
and  bottom  being  \(d-d^)\  then  I  =  ^(bd3 -Mi3).  For  the 
/  section  in  Fig.  43/  where  the  thickness  of  each  flange  is  /]  and 
that  of  the  web  is  /,  the  moment  of  inertia  /  is  also  j^(bd3  -bidi3) 
if  bi  represents  b  —  t  and  di  represents  d  —  2l\. 

The  moment  of  inertia  of  a  rectangle  with  respect  to  an  axis 
through  its  base  is  frequently  needed.  Fig.  43g  shows  this  case, 


ART.  43 


MOMENTS  OF  INERTIA 


107 


and  da  is  b  .  dz\  then  J  bz2dz  is  to  be  taken  between  the  limits 
d  and  o,  which  gives  I\  =$bd3.  This  is  seen  to  be  four  times 
as  great  as  the  moment  of  inertia  for  a  parallel  axis  through 
the  center  of  gravity. 

By  using  the  result  of  the  last  paragraph,  the  moment  of 
inertia  of  the  T  section  in  Fig.  43 h  with  respect  to  an  axis  through 
its  center  of  gravity  can  easily  be  written.  Let  the  distance 
c  and  GI  be  first  found  by  Art.  42;  let  /i  be  the  thickness  of  the 
flange  and  /  that  of  the  web.  Then  J/c3  is  the  moment  of 
inertia  of  the  portion  of  the  section  below  the  axis,  and 
%bci3  —  3(b  —  t)(ci—ti)3  is  the  moment  of  inertia  of  the  portion 
above  the  axis.  The  sum  of  these  is  the  required  moment  of 
inertia  of  the  entire  section. 


U  —  &--—,' 


t-—  6— 


__ 

—  --h 

"i 

P 

lc 

--- 

— 

.___! 

cJ 

L, 

Fig.  43h 

Fig.  43i 

Fig.  43e         Fig.  43/  Fig.  43g 

The  moment  of  inertia  of  any  section  area  is  less  for  an  axis 
through  its  center  of  gravity  than  for  any  parallel  axis.  If  / 
is  the  value  for  the  axis  through  the  center  of  gravity,  /i  that 
for  any  parallel  axis,  a  the  area  of  the  section,  and  h  the  distance 
between  the  two  axes,  the  formula  I*=Ii-ah2  furnishes  the 
means  of  finding  7  when  I\  is  known.  For  example,  take  the 
rectangle  of  breadth  b  and  depth  d  where  I\  is  known  to  be  %bd3 
for  an  axis  passing  through  one  of  the  bases;  then  for  a  parallel 
axis  through  the  center  of  gravity  of  the  rectangle,  the  value  of  7 
is  %bd3-bd(±d)2=^bd3.  Similarly,  the  value  of  7i  for  the  T  sec- 
tion in  Fig.  426  or  Fig.  43h  with  respect  to  an  axis  through  the 
base  of  the  web  is  \bd?-%(b-t)(d-ti)3,  where  d  is  the  total 
depth;  this  being  computed,  the  value  of  7  for  the  parallel  axis 
through  the  center  of  gravity  is  Ii—ac2. 

For  the  beam  of  Fig.  43z,  which  is  made  by  riveting  four 
angles  to  a  web,  the  expression  of  7  is  the  same  as  that  for  Fig.  43/. 
It  is,  however,  customary  to  use  tables  in  finding  the  numerical 
value  of  7  for  such  a  section,  these  tables  giving  the  moments 


108  GENERAL  THEORY  OF  BEAMS  CHAP,  v 

of  inertia  of  various  sizes  of  angles  with  respect  to  axes  through 
their  centers  of  gravity  and  parallel  to  the  legs.  For  example, 
let  the  horizontal  leg  of  each  angle  in  Fig.  43*  be  4  inches,  the 
vertical  leg  3  inches,  and  the  mean  thickness  |  inch.  Then 
Table  10  shows  that  the  distance  from  the  back  of  the  long  leg 
to  an  axis  through  the  center  of  gravity  of  the  angle  is  0.83  inches, 
that  its  section  area  is  3.25  square  inches  and  that  the  moment 
of  inertia  of  the  section  with  respect  to  this  axis  is  2.42  inches4. 
Four  of  these  angles  being  used  with  a  web  f  inches  thick  and 
30  inches  depth,  the  moment  of  inertia  of  the  angle  sections  with 
respect  to  the  horizontal  ax's  through  the  center. of  gravity  of 
the  web  is  4X  2.42+4x3-25(15— o. 83)2  =  2620.0  inches4,  while 
the  moment  of  inertia  of  the  web  itself  is  -^  X  f  X  3o3  =  843 . 7  inches4, 
thus  giving  7  =  2620.0  +  843.7  =  3464  inches4  for  the  moment  of 
inertia  of  the  entire  section  with  respect  to  an  axis  through  its 
center  of  gravity. 

The  moment  of  inertia  of  a  plane  area  contains  the  product 
of  four  linear  dimensions  and  is  hence  expressed  in  quadratic 
units ;  for  the  sake  of  clearness  l  inches4  '  will  be  used  to  designate 
the  unit  in  which  /  is  measured.  When  lengths  are  expressed 
in  inches,  areas  are  in  square  inches,  volumes  in  cubic  inches, 
and  moments  of  inertia  in  inches4.  The  section  factor  I/c  is  the 
product  of  three  linear  dimensions  and  is  expressed  in  '  inches3 ' ; 
since  it  is  not  a  volume  the  term  cubic  inches  would  not  be 
appropriate. 

Prob.  43#.  Consult  a  book  on  theoretic  mechanics  and  obtain  a 
proof  of  the  valuable  rule  Ii  =  I+ah2. 

Prob.  436.  For  Fig.  43&  let  the  upper  flange  be  4Xi  inches,  the 
web  be  5iXi  inches,  and  the  total  depth  be  6  inches.  Compute  the 
area  of  the  section  and  the  value  of  c,  without  using  algebraic  for- 
mulas. Compute  the  moments  of  inertia  of  flange  and  web  with  re- 
spect to  horizontal  axes  through  their  centers  of  gravity.  Transfer 
these  to  the  horizontal  axis  shown  in  the  figure  and  find  /  for  the  entire 
section  with  respect  to  that  axis. 

ART.  44.     ROLLED  BEAMS  AND  SHAPES 

Soon  after  1830,  owing  to  the  rapid  progress  of  railroad  con- 
struction, wrought  iron  began  to  come  into  use  for  railroad  rails 


ART.  44  ROLLED    BEAMS    AND    SHAPES  109 

and  bridges,  and  by  1850  rolled  beams  and  angles  were  exten- 
sively employed.  The  Bessemer  and  the  open-hearth  processes 
of  making  steel,  introduced  in  1856  and  1863,  rendered  it  possible 
to  produce  ingots  from  which  steel  rails  and  beams  could  be 
rolled,  but  it  was  not  until  about  1890  that  steel  beams  began 
seriously  to  compete  with  those  of  wrought  iron.  After  this 
date,  however,  the  use  of  wrought  iron  for  beams  and  shapes 
rapidly  declined  and  since  1900  medium  steel  has  taken  its  place 
as  a  structural  material  (Art.  25). 

Table  6,  at  the  end  of  this  volume,  gives  the  properties  of 
the  most  common  sizes  of  the  medium-steel  I  beams  rolled  iii 
the  United  States  of  America  in  1905.  The  first  column  con- 
tains all  the  depths  that  are  rolled,  but  for  each  depth  other 
sizes  than  those  given  in  the  second,  third,  and  fourth  columns 
may  be  obtained;  for  instance,  for  the  1 8-inch  beam,  sizes 
may  be  ordered  which  weigh  65  and  60  pounds  per  linear  foot, 
the  width  of  the  flanges  being  intermediate  between  6.26  and 
6.00  inches.  The  sizes  marked  with  an  asterisk  are  called 
standard,  and  they  are  always  found  in  the  market,  while  the 
other  sizes  must  generally  be  specially  ordered  and  hence  cost 
a  little  more  per  pound.  The  fifth  column  of  the  table  contains 
the  moments  of  inertia  of  the  sections  with  respect  to  an  axis  AB 
drawn  through  the  center  of  gravity  of  the  section  and  normal 
to  the  web,  as  shown  in  Fig.  44<z;  the  sixth  column  contains 
the  section  factor  I/c,  which  is  obtained  by  dividing  /  by  one-half 
of  the  depth;  the  seventh  column  will  be  explained  below.  The 
remaining  columns  refer  to  an  axis  CD  through  the  center  of 
gravity  of  the  section,  and  are  for  use  in  designing  columns 
(Art.  77).  Table  13  gives  the  same  quantities  for  the  standard 
I  sections  used  in  Germany,  the  weights  being  in  kilograms  per 
meter  of  length  and  the  linear  dimensions  in  centimeters. 

By  the  help  of  Tables  6  and  13  the  flexure  formula  (41)  is 
readily  applied  to  the  discussion  of  I  beams.  For  example,  let  a 
simple  span  30  feet  long  consist  of  a  lo-inch  I  beam  weighing 
25  pounds  per  linear  foot,  and  let  it  carry  a  uniform  load  of 
175  pounds  per  linear  foot.  The  total  uniform  load  then  is 
V  =  200  pounds  per  linear  foot,  and  by  Art.  38  the  maximum 


110 


GENERAL  THEORY  OF  BEAMS 


CHAP.  V 


bending  moment  is  M  =  ^wl2  =  22  500  pound-feet  =270  ooo  pound- 
inches.  The  horizontal  unit-stress  on  the  upper  or  lower  side 
of  the  beam  at  the  middle  of  the  span  is  now  found  from  the 
flexure  formula  to  be  S=M/(I/c)  =11  020  pounds  per  square 
inch,  this  being  tension  on  the  lower  side  and  compression  on 
the  upper  side. 

Table  7  contains  similar  quantities  for  sections  of  bulb  or 
deck  beams  like  that  shown  in  Fig.  446,  the  fifth  column  giving 
the  distances  from  the  base  of  the  flange  to  an  axis  AB  through 
the  center  of  gravity  and  normal  to  the  web;  this  distance  sub- 
tracted from  the  depth  of  the  beam  gives  the  value  of  c,  the  dis- 
tance from  the  neutral  axis  to  the  end  of  the  bulb  or  head.  Bulb 
beams  are  used  only  for  floors  where  the  beams  are  visible,  but 
I  beams  are  employed  in  all  kinds  of  floors  as  well  as  in  bridges 
and  many  other  engineering  constructions. 


Fig.  44s     Fig.  44/ 


Beams  formed  by  riveting  together  plates  and  channels, 
or  plates  and  angles,  are  in  common  use.  Tees,  channels,  angles, 
and  other  forms  which  alone  are  not  used  for  beams  are  called 
'shapes',  and  Tables  8,  9,  and  10  give  elements  of  T,  C,  and  f 
sections  which  are  of  value  in  designing  such  sections.  Many 
other  sizes  than  those  given  in  the  tables  are  found  in  the  mar- 
ket or  may  be  specially  ordered,  and  the  handbooks  of  the  manu- 
facturers contain  lengthy  tables  of  the  elements  of  such  sections. 
Fig.  44c  shows  a  tee  which  is  occasionally  used  as  a  beam,  Fig.  44d 
the  channel  section,  Fig.  440  an  angle  section  with  legs  of  equal 
length,  and  Fig.  44/  an  angle  section  with  unequal  legs.  For 
channels  and  angles  the  center  of  gravity  usually  falls  without 
the  section  as  seen  in  the  figures;  through  this  point  rectangular 
axes  are  drawn  as  shown,  and  the  tables  give  the  moment  of 
inertia  with  respect  to  each  of  these  axes. 


ART.  44 


ROLLED  BEAMS  AND  SHAPES 


111 


By  the  help  of  Table  9  the  moment  of  inertia  of  the  section 
of  the  compound  beam  shown  in  Fig.  44g  can  be  readily  com- 
puted. For  example,  let  each  channel  be  10  inches  deep  and 
weigh  15  pounds  per  linear  foot;  then  from  the  table  the  moment 
of  inertia  of  the  two  channels  with  respect  to  the  neutral  axis  ia 
2X66.9  =  133.8  inches4.  Let  each  plate  be  9X1  inches  in  sec> 
tion;  then  by  Art.  43  the  moment  of  inertia  of  each  with  respect 
to  a  horizontal  axis  through  its  center  of  gravity  is  -rjXgXi3  = 
0.75  inches4,  and  the  moment  of  inertia  of  the  two  plates  with 
respect  to  the  neutral  axis  of  the  beam  is  2X0. 75  +  2X9.00X5. 52  = 
'46.0  inches4.  Accordingly  the  required  moment  of  inertia  of 
the  section  is  1=133.8  +  546.0  =  679.8  inches4,  and  the  section 
factor  is  I/c=  679.8/6.0=  113.3  inches3. 

1* — 6 — H 
\ 


Fig.  44g        Fig.  44h  Fig.  44*  Fig.  44;  Fig.  44k 

The  complete  theory  of  moment  of  inertia  is  a  lengthy  one 
and  properly  belongs  to  pure  mathematics;  see  Art.  170  forfullei 
discussion.  It  may  be  noted  here,  however,  that  the  sum  of 
the  moments  of  inertia  of  a  plane  area  with  respect  to  two 
rectangular  axes  through  a  given  point  is  always  a  constant 
whatever  be  the  directions  of  those  axes.  Thus,  for  the 
rectangle  in  Fig.  444,  the  moment  of  inertia  with  respect  to  a  hori- 
zontal axis  through  the  center  of  gravity  by  Art.  43,  is  -fabd? 
and  that  for  a  vertical  axis  through  the  same  point  is  T?db3.  If 
any  other  rectangular  axes  be  drawn,  as  in  the  figure,  the  moments 
of  inertia  with  respect  to  them  are  less  than  -fybd3  and  greater 
than  -f^db3,  but  their  sum  is  equal  to  ^bd(d2  +  b2).  In  this  case 
the  greatest  moment  of  inertia  for  an  axis  through  the  center 
of  gravity  is  when  the  axis  is  parallel  to  the  short  side  of  the 
rectangle,  and  the  least  is  when  the  axis  is  parallel  to  the  long 
side.  Similarly  the  moments  of  inertia  for  the  channel  sections  in 
Table  9  are  the  greatest  and  least  with  respect  to  axes  through  the 
center  of  gravity  which  are  perpendicular  and  parallel  to  the  web. 


112  GENERAL  THEORY  OF  BEAMS  CHAP.  V 

For  the  angle  sections  in  Table  10  the  given  moments  of 
inertia  are  with  respect  to  axes  through  the  center  of  gravity 
and  parallel  to  the  legs,  as  shown  in  Figs.  44e  and  44/.  These 
are  the  values  usually  required  in  designing,  but  it  may  be  men- 
tioned that  they  are  not  the  greatest  and  least  values.  Fig.  44; 
shows  the  approximate  position  of  the  two  rectangular  axes 
for  which  the  moments  of  inertia  are  greatest  and  least,  MN 
giving  the  greatest  value  and  mn  the  least.  When  the  legs  of 
the  angles  are  equal,  each  of  these  axes  is  inclined  to  the  back 
of  the  legs  at  an  angle  of  45  degrees. 

The  center  of  gyration  of  a  section  is  the  point  where  the 
entire  area  might  be  concentrated  and  have  the  same  moment 
of  inertia  as  the  actual  distributed  area.  The  distance  from 
the  axis  to  this  center  is  called  the  'radius  of  gyration  '  and  this 
is  designated  by  r.  From  this  definition  it  follows  that  I  =  ar2, 
or  r2=I/a.  Values  of  the  radius  of  gyration  for  rolled  steel 
sections  with  respect  to  axes  through  their  centers  of  gravity 
are  given  in  Tables  6-10.  Fig.  44k  indicates  that  the  distance  r  is 
always  less  than  the  distance  c  from  the  neutral  axis  to  the 
remotest  fiber. 

Prob.  44a.  Show  that  the  weight  of  the  medium-steel  beams  and 
shapes  in  Tables  6-10  is  3.4  pounds  per  linear  foot  for  each  square  inch 
of  cross-section,  or  489.6  pounds  per  cubic  foot. 

Prob.  446.  Let  the  section  in  Fig.  44h  consist  of  a  plate  JX  24 
inches  and  four  angles  each  4X3X3-  inches  in  size,  the  longer  leg  of 
the  angle  being  horizontal.  Compute  the  moment  of  inertia  of  the 
section  with  respect  to  its  neutral  axis. 

ART.  45.     ELASTIC  DEFLECTIONS 

When  a  beam  is  subject  to  the  action  of  loads  the  horizontal 
fibers  on  one  side  of  the  neutral  axis  are  elongated  and  those 
on  the  other  side  are  shortened  (Art.  40).  The  beam  therefore 
bends  and  all  points  except  those  over  the  supports  deflect  below 
their  original  position.  The  curve  assumed  by  the  neutral  sur- 
face of  the  beam,  when  the  elastic  limit  of  the  material  is  not 
exceeded,  is  called  the  'Elastic  Curve',  and  its  general  equation 
will  now  be  deduced. 


ART.  45 


ELASTIC  DEFLECTIONS 


Let  mn  in  Fig.  45a  or  456  be  any  short  length  measured  along 
the  neutral  surface  of  the  beam.  Let  kk'  and  pp'  be  two  normal 
sections  passing  through  m  and  n\  before  the  bending  kV  and  pp' 
were  parallel,  after  the  bending  they  intersect  at  o,  the  center 
of  curvature.  Let  qqf  be  drawn  through  n  parallel  to  kkf.  Then 
qp  represents  the  elongation  of  the  fiber  kq,  and  <ftf  the  shortening 
of  the  fiber  k'q',  and  these  changes  of  length  are  proportional 
to  their  distances  from  the  neutral  surface.  Let  the  change 
of  length  qp  be  called  e  and  let  61  be  the  corresponding  unit- 
stress  on  the  fiber  kp,  and  E  be  the  modulus  of  elasticity  of  the 
material.  Let  the  length  of  the  beam  be  /  and  the  short  dis- 
tance mn  be  dl.  Then,  by  Art.  9,  the  value  of  qp  is  e  =•=  (S/E)dl. 


°! 


Fig.  45a 


Fig.  456 


Now  let  c  be  the  distance  from  qp  to  the  neutral  surface  of 
the  beam,  and  let  R  be  the  radius  of  curvature  of  the  elastic 
curve.  From  the  similar  figures  omn  and  pnq,  it  follows  that 

om/mn=nq/qp  or  $L/dl=c/e 

and,  inserting  for  e  its  value  (S/E)  dl,  this  becomes  S/c  =  E/~R, 
But  the  flexure  formula  (41)  gives  S/c  =  M/I,  where  M  is  the 
bending  moment  of  the  external  forces  and  /  is  the  moment  of 
inertia  of  the  section  area.  Accordingly, 


or 


which  gives  the  relation  between  the  radius  of  curvature  of  the 
elastic  curve  at  any  section  and  the  bending  moment  at  that 
section.  When  there  is  no  bending  moment,  R  is  infinity  or 
the  curve  is  a  straight  line;  where  M  has  its  greatest  value,  there 
R  has  its  least  value  and  the  curvature  is  the  sharpest. 

Now  the  radius  of  curvature  of  any  plane  curve  for  which 
?he  abscissa  is  x,  the  ordinate  y,  and  the  length  /,  is  ascertained 


114  GENERAL  THEORY  OF  BEAMS  CHAP.  V 

in  works  on  differential  calculus  to  be  given  by  R=dl3/dx .  d2y. 
If  this  value  of  R  be  equated  to  EI/M,  there  results  a  general 
differential  equation  of  the  elastic  curve  which  applies  to  the 
flexure  of  all  beams  or  arches  in  which  the  elastic  limit  of  the  mate- 
rial is  not  exceeded.  In  discussing  a  beam  the  axis  of  x  is  taken 
as  horizontal  and  that  of  y  as  vertical.  Since  experience  teaches 
that  the  length  of  a  small  part  of  a  bent  beam  does  not  mate- 
rially differ  from  that  of  its  horizontal  projection,  the  length  dl 
may  be  placed  equal  to  dx,  and, 

d2v    M  d2v 

B-B      or      EIU=M  (45) 

which  is  the  differential  equation  for  the  discussion  of  the  elastic 
deflection  of  beams.  In  this  formula  y  is  the  ordinate  of  the 
elastic  curve  at  the  point  where  the  abscissa  is  xt  and  the  rela- 
tion between  y  and  x  is  to  be  obtained  by  integrating  the  differen- 
tial equation  twice  and  determining  the  constants  of  integration; 
for  this  purpose  M  is  to  be  expressed  as  a  function  of  x.  Unless 
otherwise  stated  /  will  be  regarded  as  a  constant,  that  is,  the 
beam  is  of  uniform  section  throughout  its  length. 


Numerous  applications  of  this  general  formula  will  be  given 
in  the  following  chapters,  and  only  the  simple  case  of  a  canti- 
lever beam  under  a  load  at  the  end  will  be  here  discussed.  Let 
Fig.  45c  represent  the  beam  which  is  horizontal  at  the  wall,  / 
being  its  length.  Let  the  origin  of  coordinates  be  taken,  at  the 
wall,  values  of  x  being  positive  toward  the  left  and  upward  values 
of  y  positive.  The  bending  moment  at  any  section  distant  l  —  x 
from  the  left  end  is  -P(l-x),  and  the  differential  equation  (45) 
for  this  case  is, 


Integrating  this  equation,  and  determining  the  constant  of  in- 
tegration by  the  condition  that  the  value  of  the  tangent  dyjdx 


ART.  45  ELASTIC    DEFLECTIONS  115 

becomes  zero  when  x  equals  zero  there  results, 


Integrating  again  and  determining  the  constant  by  the  condi- 
tion that  y  equals  zero  when  x  is  zero,  there  is  found, 


which  is  the  equation  of  the  elastic  curve.    When  x  equals  I 
the  value  of  y  is  the  deflection  of  the  end,  which  is  —PI3/  3  EL 

The  equation  M  R  =  El  shows  that  M  and  R  must  always  have 
the  same  sign,  since  El  can  never  be  negative.  When  the  bend- 
ing moment  M  is  positive,  R  is  also  positive  and  it  is  directed 
upward  as  in  Fig.  450;  when  M  is  negative,  R  is  also  negative 
and  it  is  directed  downward  as  in  Fig.  456.  A  positive  bending 
moment  hence  indicates  that  the  lower  side  of  the  beam  is  con- 
vex or  in  tension,  while  a  negative  bending  moment  indicates 
that  the  lower  side  of  the  beam  is  concave  or  in  compression. 

Prob.  450.  A  straight  stick  of  wood,  2X2X18  inches,  is  laid  on 
two  supports  very  near  its  ends.  Compute  the  radius  of  curvature  of 
the  elastic  curve  at  the  middle. 

Prob.  45&.  Prove,  when  two  equal  loads  on  a  simple  beam  are 
equally  distant  from  the  middle,  that  the  elastic  curve  of  the  part  of 
the  beam  between  them  is  a  circle. 

Prob.  45^.  Find  the  equation  of  the  elastic  curve  for  the  cantilever 
beam  of  Fig.  45^  under  uniform  load,  taking  the  origin  of  coordinates 
at  the  free  end  and  letting  values  of  x  be  positive  toward  the  right. 


116  CANTILEVER  AND  SIMPLE  BEAMS  CHAP.  VI 

CHAPTER  VI 
CANTILEVER    AND    SIMPLE    BEAMS 

ART.  46.     SHEAR  AND  MOMENT  DIAGRAMS 

The  fundamental  principles  relating  to  beams  have  been 
deduced  in  the  preceding  chapter  and  it  now  only  remains  to 
apply  them  to  special  cases.  The  shear  formula  S8 .  a  =  V  and 
the  flexure  formula  S.I/c  =  M  (Art.  41)  contain  all  the  quan- 
tities needed  for  investigation  or  design,  and  the  first  step  is 
to  obtain  the  values  of  the  vertical  shear  V  and  the  bending 
moment  M\  for  brevity  these  will  hereafter  be  called  '  shear  ' 
and  'moment '.  Arts.  37  and  38  show  that  V  and  M  vary  through- 
out the  beam,  and  the  manner  in  which  they  vary  will  now  be 
further  discussed.  The  maximum  values  of  F  and  M  will  give 
the  greatest  values  of  the  unit-stresses  S8  and  S. 

The  method  of  computing  the  shear  F  and  the  moment  M 
for  any  given  section  of  a  simple  beam  is  by  the  use  of  the  defi- 
nitions of  these  quantities,  namely, 

F=Left  reaction  minus  loads  on  left  of  section 
M= Moment  of  left  reaction  minus  moment  of  loads  on  left 

and  these  apply  also  to  cantilever  beams  by  making  the  left 
reaction  equal  to  zero,  since  there  is  no  reaction  at  the  left  end. 
From  these  definitions,  values  of  F  and  M  for  several  sections 
are  readily  computed  and  graphical  representations  of  the  results 
may  then  be  made. 

The  following  figures  show  shear  and  moment  diagrams  for 
a  cantilever  beam,  Fig.  46a  being  for  a  uniform  load,  Fig.  466 
for  a  concentrated  load  at  the  free  end,  and  Fig.  46c  for  both 
uniform  and  concentrated  load,  the  upper  diagram  being  for 
the  shears  and  the  lower  one  for  the  moments.  To  explain 
the  manner  of  their  construction,  it  will  be  sufficient  to  consider 
only  the  third  case.  Let  the  beam  be  15  feet  long,  the  uniform 


ART.  46  SHEAR    AND    MOMENT    DIAGRAMS  117 

load  be  40  pounds  per  linear  foot  and  the  concentrated  load  be 
700  pounds.  The  shear  at  the  free  end  on  the  right  of  the  con- 
centrated load  is  V  =  —  700  pounds,  at  5  feet  from  the  left  end 
it  is  V=  -700-5X40=  -900  pounds,  at  10  feet  from  the  left 
end  it  is  F=  —700  —  10X40=  —  i  100  pounds,  and  at  the  wall 
it  is  V=  —700  —  15X40=.—  i  300  pounds;  these  values  are  laid 
off  to  scale  downwards  from  a  horizontal  line  and  the  line  con- 
necting their  lower  ends  is  then  drawn.  The  moment  at  the 
free  end  is  zero,  since  the  lever  arm  of  the  concentrated  load 
for  this  section  is  zero,  the  moment  at  5  feet  from  the  left  end 
is  M  =  —  700X5  —  200X2 J=  —  4  ooo  pound-feet,  at  10  feet  from 
the  left  end  it  is  M=  —700X10  —  400X5  =  —9  ooo  pound-feet, 
and  at  the  wall  it  is  M=  —700X15—  600 Xj^=  —  15  ooo  pound- 
feet  ;  these  values  are  laid  off  to  scale  and  a  line  connecting  their 
lower  ends  is  drawn.  The  diagrams  thus  constructed  shows 
clearly  the  distribution  of  the  shears  and  moments  throughout 
the  beam. 


Fig.  46a  Fig.  466  Fig.  46c 

The  following  figures  show  shear  and  moment  diagrams  for 
a  simple  beam,  Fig.  46d  being  for  uniform  load,  Fig.  46e  for 
one  concentrated  load,  and  Fig.  46/  for  both  uniform  and  con- 
centrated loads.  For  the  last  case  let  the  beam  be  18  feet  long  and 
weigh  20  pounds  per  linear  foot,  the  larger  load  being  360  pounds 
at  6  feet  from  the  left  end  and  the  smaller  one  being  90  pounds 
at  12  feet  from  the  left  end.  The  reactions  of  the  left  and  right 
supports  are  found  by  Art.  36  to  be  450  and  360  pounds,  the 
sum  of  which  equals  the  total  load,  810  pounds.  To  construct 
the  shear  diagram  a  sufficient  number  of  values  of  V  must  be 
computed;  thus,  at  the  right  of  the  left  support  F=+45o,  at 
the  left  of  the  first  load  V=  +450  — 120  = +330,  at  the  right  of 
that  load  V=  +450-120-360  =  -30,  at  the  middle  of  the 


118  CANTILEVER  AND  SIMPLE  BEAMS  CHAP,  vi 

span  7= +450-180-360= —90  pounds,  and  so  on.  To  con- 
struct the  moment  diagram  several  values  of  M  are  computed; 
thus  at  3  feet  from  the  left  end  M  =  450X3—  6oXiJ=  +i  260, 
under  the  first  load  ^=450X6  —  120X3  =  +2  340,  at  the  middle 
of  the  span  M  =450X9-180X4^  —  360X3=  +2  160  pound- 
feet,  and  so  on.  These  values  being  laid  off  as  ordinates  and 
lines  being  drawn  connecting  their  ends,  the  distribution  of 
the  shears  and  moments  throughout  the  beam  is  graphically 
represented. 


Fig.  4Gd  Fig.  46e  Fig.  46/ 

From  the  definitions  of  V  and  M  and  from  the  general  dis- 
cussions in  Arts.  37  and  38,  it  is  seen  that  the  shears  are  always 
represented  by  ordinates  to  straight  lines  which  are  inclined 
when  uniform  load  is  considered  and  horizontal  when  concen- 
trated loads  are  alone  considered.  It  is  also  seen  that  the  moments 
are  represented  by  ordinates  to  straight  lines  for  concentrated 
loads  alone  and  by  ordinates  to  parabolic  curves  for  uniform 
load  either  alone  or  accompanied  by  concentrated  loads. 

The  above  diagrams  for  simple  beams  show  that  the  maxi- 
mum moment  occurs  at  the  section  where  the  shear  passes  through 
zero.  This  is  indeed  a  general  law  the  truth  of  which  will  be 
demonstrated  in  the  next  article.  This  law  also  applies  to  canti- 
lever beams,  for  at  the  wall  there  is  a  reaction  equal  to  the  weight 
of  the  beam  and  load,  hence  on  passing  that  section  the  shear 
becomes  zero. 

Prob.  460.  Construct  shear  and  moment  diagrams  for  a  cantilevei 
beam  10  feet  long  and  weighing  13  pounds  per  linear  foot,  there  being 
a  concentrated  load  of  75  pounds  at  2  feet  from  the  left  end. 

Prob.  466.  Construct  shear  and  moment  diagrams  for  a  simple 
beam  20  feet  long  and  weighing  13  pounds  per  linear  foot,  there  being 
a  concentrated  load  of  240  pounds  at  5  feet  from  the  left  end. 


AJIT.  47  MAXIMUM  SHEARS  AND  MOMENTS  119 


ART.  47.     MAXIMUM  SHEARS  AND  MOMENTS 

The  greatest  numerical  value  of  the  shear  or  moment  which 
occurs  in  a  beam  under  a  given  system  of  loads  is  called  the 
maximum  value,  whether  it  be  positive  or  negative.  These 
maximum  values  are  to  be  used  in  the  shear  and  flexure  formulas 
of  Art.  41  without  regard  to  sign,  for  the  signs  +  and  —  pre- 
fixed to  a  shear  indicate  merely  whether  the  left  part  of  the  beam 
tends  to  slip  up  or  down  with  respect  to  the  other  part  (Art.  37), 
while  when  prefixed  to  a  moment  they  indicate  merely  whether 
the  upper  side  of  the  beam  is  concave  or  convex  (Art.  44). 

For  a  cantilever  beam  both  maximum  shear  and  moment 
occur  at  the  wall.  Let  w  be  the  uniform  load  per  linear  unit, 
/  the  length  of  the  beam,  and  W  the  total  load  wl\  then  the  max- 
imum shear  is  V=—wl=—W,  and  the  maximum  moment  is 
M=  —  WX\l  =  —%Wl.  Let  P  be  any  concentrated  load  at  the 
left  end  of  the  beam,  as  in  Fig.  466,  then  the  maximum  shear 
is  V  =  —  P  and  the  maximum  moment  is  M=  —PI.  Or,  without 
regard  to  sign,  the  maximums  for  a  cantilever  beam  are, 

For  uniform  load  V=W  M=$Wl 

For  Pat  the  end  V=P  M=Pl 

1 1  P  is  not  at  the  free  end  but  at  the  distance  p  from  that  end, 
as  in  Fig.  47a,  the  maximum  shear  is  also  P,  but  the  maximum 
moment  is  P(l  —  p). 

L/~  '«- P--JP 


-*|  ^  % X 1 

Fig.  47a  Fig.  476 

For  a  simple  beam  under  uniform  load  alone,  each  reaction 
is  \wl  and  the  shear  at  any  section  distant  x  from  the  left  support 
is  V=  +%wl-u>x;  the  maximum  shears  hence  occur  when 
#  =  o  and  x  =  l,  their  values  being  +\W  and  -\W .  The 
moment  for  the  section  distant  x  from  the  left  support  is 
M=  +  \wl .  x-wx  .  \xy  and  by  the  usual  method  of  the  dif- 
ferential calculus  this  is  found  to  be  a  maximum  when  #  =  J/, 


120  CANTILEVER  AND  SIMPLE  BEAMS  CHAP.  VI 

and  for  this  value  M  =  +  Jw/2  =  %Wl.  For  a  single  load  P  at 
the  middle  of  a  simple  beam,  each  reaction  is  JP  and  this  is  the 
maximum  shear;  also  the  moment  at  any  section  between  the 
left  support  and  the  middle  is  M  =  +  %Px,  and  this  is  a  maximum 
when  x=$l,  as  Fig.  46e  shows.  Hence  without  regard  to  sign, 
the  maximums  for  a  simple  beam  are, 

For  uniform  load  V=  \W  M=\Wl 

For  P  at  the  middle  V=  JP  M=\Pl 

If  P  is  not  at  the  middle  but  at  the  distance  p  from  the  left  end, 
as  in  Fig.  476,  the  left  reaction  is  P(l  —  p)/l  and  the  right  reaction 
is  Pp/l  and  these  are  the  maximum  shears;  the  maximum 
moment  is  under  the  load  and  its  value  is  Rip  or  P(l  —  p)p/l, 
which  has  its  greatest  value  \Pl  when  p  equals  J/. 

When  one  or  more  concentrated  loads  are  on  a  simple  beam, 
the  maximum  shears  are  the  reactions  of  the  supports,  and  the 
maximum  moment  due  to  these  .loads  occurs  under  one  of  the 
loads,  as  in  Fig.  38c.  The  maximum  moment  due  to  both 
uniform  and  concentrated  loads  also'  often  occurs  under  one  of 
the  single  loads,  as  seen  in  Fig.  46/,  but  it  sometimes  occurs 
at  a  different  section,  as  is  the  case  in  Prob.  463.  The  section 
where  the  moment  is  a  maximum  is  called  the  'dangerous  sec- 
tion ',  since  there  the  greatest  horizontal  stress  S  will  be  found 
by  the  use  of  the  flexure  formula  S.I/c  =  M  (Art.  41).  In 
order  to  locate  the  dangerous  section,  the  following  important 
law  may  be  used. 

The  dangerous  section  is  that  where  the  shear  passes  through 
zero.  To  prove  this,  let  there  be  several  loads  on  a  simple  beam, 
PI  at  the  distance  pi  and  PI  at  the  distance  p2  from  the  left 
support,  and  so  on.  Let  RI  be  the  left  reaction  due  to  these  loads 
and  the  uniform  load  wl.  Then  the  bending  moment  for  a 
section  distant  x  from  the  left  support  is, 


in  which  only  the  single  loads  appear  which  are  between  the 
left  support  and  the  section.  The  value  of  x  which  makes 
the  bending  moment  a  maximum  is  obtained  by  equating  to 


ART.  48  INVESTIGATION    OF    BEAMS  121 

zero  the  derivative  of  M  with  respect  to  xt  or 
8M/dx=Ri-wx-Pl-P2-etc.=o 

is  the  equation  which  gives  the  value  of  x.  But  RI  -wx  —  P\  —P2 
is  the  shear  V  for  this  section,  as  is  clear  from  the  definition  of 
vertical  shear  in  Art.  37.  Therefore  the  maximum  moment 
occurs  at  the  section  where  the  shear  passes  through  zero.  This 
section  can  readily  be  found  by  computing  shears  for  different 
sections,  and  the  construction  of  the  shear  diagram  will  always 
be  of  assistance. 

For  example  take  the  data  of  Prob.  466  where  I  =  20  feet, 
w=  13  pounds  per  linear  foot,  and  PI  =  240  pounds  at  pi  =5  feet. 
The  left  reaction  ^1=130  +  1X240=310  pounds.  The  shear 
just  at  the  left  of  P  is  +310-65  =  +245  and  just  at  the  right 
of  P  it  is  +310  —  65  —  240=  +5  pounds,  and  hence  V  does  not 
pass  through  zero  under  the  single  load.  To  find  the  exact 
position  of  the  dangerous  section,  let  x  be  its  distance  from  the 
left  support;  then  the  shear  is  310-13^-240  and  equating 
this  to  zero,  there  is  found  x=  5.385  feet.  The  maximum  moment 
is  310 X  5.385  -  i  X 13  X  5.3852  -  240 Xo.385  =  i  388  pound-feet. 

Prob.  470.  Compute  the  maximum  shear  and  moment  for  a  canti- 
lever beam  13  feet  long  which  weighs  33  pounds  per  linear  foot  and 
has  a  single  load  of  375  pounds  at  3  feet  from  the  free  end. 

Prob.  476.  A  simple  beam  of  12  feet  span  weighs  35  pounds  per 
linear  foot  and  has  three  concentrated  loads  of  300,  60,  and  150  pound? 
at  3,  5,  and  8  feet  respectively  from  the  right  support.  Compute  the 
maximum  shear  and  moment,  and  draw  the  shear  and  mpmenf 
diagrams. 

ART.  48.     INVESTIGATION  OF  BEAMS 

The  investigation  of  a  beam  of  constant  cross-section  usually 
consists  in  computing  the  greatest  horizontal  unit-stress  S  from 
the  flexure  formula  S  .  I/c  =M  which  was  established  in  Art.  41, 
For  this  purpose  the  forrmTa  may  be  written 

S= Me  / 1  or  S=M/(I/c)  (48) 

the  first  of  which  may  be  used  when  c  is  determined  by  Art.  42 


122  CANTILEVER  AND  SIMPLE  BEAMS  CHAP.  VI 

and  /  by  Art.  43,  and  the  second  when  the  section  factor  I/c 
for  rolled  beams  is  taken  from  tables  (Art.  44).  The  greatest 
value  of  S  will  be  found  at  the  dangerous  section  where  If  is  a 
maximum  and  hence  the  maximum  moment  is  to  be  ascertained 
from  Art.  47.  If  M  is  computed  in  pound-feet,  it  must  be 
reduced  to  pound-inches  when  /  and  c  are  expressed  in  terms 
of  inches;  then  the  value  of  S  will  be  in  pounds  per  square  inch. 

The  unit-stress  S  will  be  tension  or  compression  according 
as  the  remotest  fiber  from  the  neutral  surface  lies  on  the  convex 
or  concave  side  of  the  beam,  c  being  the  distance  between  that 
fiber  and  the  neutral  surface.  If  S'  is  the  unit-stress  on  the 
opposite  side  of  the  beam  and  cr  the  distance  from  it  to  the  neutral 
surface,  then  by  Art.  40, 

S/c=S'/c'  or  S'=S.c'/c 

When  S  is  tension,  S'  is  compression;  when  S  is  compression, 
S'  is  tension.  Sometimes  it  is  necessary  to  compute  S'  as  well 
as  S  in  order  thoroughly  to  investigate  the  stability  of  the  beam. 
By  comparing  the  values  of  S  and  S'  with  the  proper  working 
unit-stresses  for  the  given  materials  (Art.  7),  the  degree  of  security 
of  the  beam  may  be  inferred. 

As  an  example  consider  a  cast-iron  I  beam  which  has  a  depth 
of  10  inches,  width  of  flanges  6  inches,  thickness  of  flanges  and 
web  i  inch.  It  is  supported  at  its  ends  forming  a  span  of  12 
feet,  and  carries  two  loads,  each  weighing  5  ooo  pounds,  one 
being  at  the  middle  and  the  other  at  one  foot  from  the  right  end. 
The  steps  of  the  computation  are  as  follows: 
by  geometry,  a=  20  square  inches 

by  Art.  42,  c=     5  inches 

by  Art.  43,  /=  286.7  inches4 

by  Art.  17,  w=  62.7  pounds  per  linear  foot 

by  Art.  47,  x=     6  feet  for  dangerous  section 

by  Art.  38,          max.  M=iS  630  pound-feet 

Then  from  the  flexure  formula,  the  unit-stress  is, 

5=i8  600X12X5/287  =  3  900  pounds  per  square  inch 
which  is  the  tensile  stress  on  the  lower  side  of  the  beam  and 


ART.  48  INVESTIGATION    OF    BEAMS  123 

the  compress! ve  stress  on  the  upper  side.  The  factor  of  safety 
in  compression  is  90  ooo/  3  900  or  about  23,  while  that  in  tension 
is  20  000/3  900  or  about  5 ;  the  degree  of  security  for  compres- 
sion is  ample,  but  that  for  tension  is  too  low  (Art.  7). 

As  a  second  example,  consider  a  simple  wooden  beam,  3 
inches  wide,  4  inches  deep,  and  16  feet  span,  and  let  a  man  weigh- 
ing 150  pounds  stand  at  the  middle.  Here  6=3  inches,  d  = 
4  inches,  /  =  i6  feet  =  192  inches,  c=\d  =  2  inches,  I=^bd3  = 
16  inches4,  the  weight  of  the  beam  ^  =  10X12X5^/12=53.3 
pounds,  and  P  =  i5o  pounds.  The  dangerous  section  is  at  the 
middle,  and  the  maximum  moment  due  to  the  uniform  and 
single  loads  is,  from  Art.  47,  M  =\Wl  +  ^Pl  =  jo6.j  pound-feet  = 
8  480  pound-inches.  Then  the  flexure  formula  gives  S  = 
8480X2/16  =  1060  pounds  per  square  inch,  which  is  a  satis- 
factory unit-stress  for  the  tension  of  timber  under  a  steady  load, 
but  is  a  little  too  high  for  compression. 

A  short  beam  heavily  loaded  should  also  be  investigated  for 
shearing  at  the  supports  by  the  shear  formula  S8a  =  V  (Art.  41), 
but  for  common  cases  there  is  ample  security  against  this  stress. 
For  the  cast-iron  beam  above  discussed,  the  maximum  shear 
is  at  the  right  support  and  its  value  is  7  460  pounds ;  hence 
£3  =  7460/20=373  pounds  per  square  inch,  so  that  the  factor 
of  safety  against  shearing  is  about  48.  Similarly  for  the  timber 
beam,  the  factor  of  safety  against  shearing  may  be  found  to  be 
greater  than  350. 

When  the  load  upon  a  beam  is  heavy  compared  with  its  own 
weight,  the  latter  may  be  omitted  from  the  computation  as  its 
influence  is  small.  A  common  rule  in  practice  is  that  the  weight 
of  the  beam  may  be  neglected  whenever  the  moment  due  to 
it  is  less  than  ten  percent  of  that  due  to  the  loads  on  the  beam; 
for  a  concentrated  load  at  the  middle  this  will  be  the  case  when 
\Wl  is  less  than  one- tenth  of  \Pl,  that  is,  when  P  is  greater 
than  $W. 

Prob.  480.  A  piece  of  scantling  2  inches  square  and  10  feet  long  is 
hung  horizontally  by  a  rope  at  each  end  and  two  students  stand  upon 
it.  Is  it  safe  ? 


124  CANTILEVER  AND  SIMPLE  BEAMS  CHAP.  VI 

Prob.  486.  A  cast-iron  bar  one  inch  in  diameter  and  two  feet  long 
is  supported  at  its  middle  and  a  load  of  200  pounds  hung  upon  each 
end  of  it.  Find  its  factor  of  safety. 


ART.  49.    SAFE  LOADS  FOR  BEAMS 

The  proper  load  for  a  beam  should  not  make  the  value  of  S 
at  the  dangerous  section  greater  than  the  allowable  unit-stress. 
This  allowable  unit-stress  or  working  strength  may  be  assumed 
according  to  the  circumstances  of  the  case  by  first  selecting  a 
suitable  factor  of  safety  from  Art.  7  and  dividing  the  ultimate 
strength  of  the  material  by  it,  the  least  ultimate  strength  whether 
tensile  or  compressive  being  taken.  For  any  given  beam  the 
quantities  /  and  c  are  known.  Then,  in  the  flexure  formula 
M  =S  .  I/c  the  maximum  moment  M  may  be  expressed  in  terms 
of  the  length  of  the  beam  and  the  unknown  loads,  and  thus  those 
loads  be  found. 

As  an  example,  consider  a  wooden  cantilever  beam  whose 
length  is  6  feet,  breadth  2  inches,  depth  3  inches,  and  which  is 
loaded  uniformly  with  w  pounds  per  linear  foot.  It  is  required 
to  find  the  value  of  w  so  that  S  may  be  800  pounds  per  square 
inch.  Here  c  =  i%  inches,  /  =TT>  and  M=^6x6w.  Then,  from 
the  flexure  -formula,  216^  =  800X54/1^X12,  whence  w  =  n 
pounds  per  linear  foot.  Since  this  wooden  beam  weighs  about 
2  pounds  per  foot,  the  total  safe  uniform  load  will  be  about 
9X6  =  54  pounds. 

As  a  second  example,  take  a  rolled  steel  I  beam  of  18  feet 
span  which  is  10  inches  deep  and  weighs  25  pounds  per  foot, 
and  let  it  be  required  to  find  what  concentrated  load  P  may 
be  put  at  the  middle  in  order  that  the  unit-stress  at  the  dangerous 
section  shall  be  15  ooo  pounds  per  square  inch.  From  Table  6 
the  value  of  the  section  factor  I/c  is  24.4  inches3.  From  Art.  47 
the  maximum  moment  is  M  =  JPXi8Xi2  pound-inches  if  the 
weight  of  the  beam  be  disregarded.  Accordingly  $^P=S  .I/c 
or  54P  =  15  000X24.4,  whence  P=6  780  pounds.  As  this  is  more 
than  five  times  the  weight  of  the  beam,  it  may  be  taken  as  the 
allowable  load  (Art.  48).  If  the  weight  of  the  beam  is  considered, 


ART.  so  DESIGNING  OF  BEAMS  125 

however,  the  moment  is  M  =  54P-fi2  150,  and  placing  this 
equal  to  S .  I/c  there  is  found  P  =  6  560  pounds,  which  is  only 
about  three  percent  less  than  the  value  obtained  before. 

As  an  example  of  an  unsymmetric  section,  let  it  be  required 
to  determine  the  total  uniform  load  W  for  a  cast-iron  T  beam  of 

14  feet  span,  so  that  the  factor  of  safety  may  be  6,  the  depth 
of  the  beam  being  18  inches,  the  width  of  the  flange  12  inches, 
the  thickness  of  the  stem  i  inch,  and  the  thickness  of  the  flange 
ij  inches.     First,  from  Art.  42  the  value  of  c  is  found  to  be 
12.63  inches,  and  that  of  c'  to  be  5.37  inches.     From  Art.  43 
the  value  of  /  is  computed  to  be  i  031  inches4.     From  Art.  47 
the   maximum  moment  is  M  =  %Wl  =  2iW  pound-inches.     Now 
with  a  factor  of  safety  of  6,  the  working  unit-stress  S  on  the 
compressive  side  of  the  dangerous  section  is  to  be  1X90000  = 

15  ooo  pounds  per  square  inch;    then,  inserting  the  values  in  the 
flexure  formula  M  =  SI/c,  the  load  W  is  found  to  be   58  300 
pounds.     Again  with  a  factor  of  safety  of  6,  the  working  unit- 
stress  S'  on  the  tensile  side  of  the  dangerous  section  is  to  be 
J  X  20  ooo  =  3  330  pounds  per  square  inch ;    inserting  the  values 
in  the  flexure  formula  M  =  SfI/c',  the  load  W  is  found  to  be 
30  400  pounds.     The  total  uniform  load  on  the  beam  should 
hence  not  exceed  30  400  pounds,  and  under  this  load  the  factor 
of  safety  on  the  compressive  side  is  nearly  12. 

Prob.  49a.  A  simple  wooden  beam,  8  inches  wide,  9  inches  deep, 
and  14  feet  in  span,  carries  two  equal  loads,  one  being  2.5  feet 
on  the  left  of  the  middle  and  the  other  2.5  feet  on  the  right.  Find 
these  loads  so  that  the  factor  of  safety  of  the  beam  shall  be  10. 

Prob.  49&.  A  steel  railroad  rail  of  2  feet  span  carries  a  load  P  at 
the  middle.  If  its  weight  per  yard  is  56  pounds,  7=12.9  inches4  and 
c=2.i6  inches,  find  P  so  that  the  greatest  horizontal  unit-stress  at  the 
dangerous  section  shall  be  6  ooo  pounds  per  square  inch. 

ART.  50.    DESIGNING  OF  BEAMS 

When  a  beam  is  to  be  designed,  the  loads  to  which  it  is  to  be 
subjected  are  known,  as  also  is  its  length,  and  from  these  the 
maximum  bending  moment  may  be  found  by  Art.  47.  The 


126  CANTILEVER  AND  SIMPLE  BEAMS  CHAP.  VI 

allowable  working  unit-stress  S  is  assumed  in  accordance  with 
engineering  practice.  Then  the  flexure  formula  (41)  may  be 

Written  I/c-M/S  (50) 

and  the  numerical  value  of  the  second  member  be  found.  The 
dimensions  to  be  chosen  for  the  beam  must  be  such  that  the 
section  factor  I/c  shall  be  equal  to  this  numerical  value,  and 
these  in  general  are  determined  by  trial,  certain  proportions 
being  first  assumed.  The  selection  of  the  proper  proportions 
and  shapes  of  beams  for  different  cases  requires  much  judgment 
and  experience;  but  whatever  forms  be  selected,  they  must  in 
each  case  be  such  as  to  satisfy  the  above  equation. 

For  instance,  a  simple  beam  of  structural  steel,  16  feet  in 
span,  is  required  to  carry  a  rolling  load  of  500  pounds.  Here, 
by  Art.  47,  the  value  of  maximum  M  due  to  the  load  of  500 
pounds  is  24  ooo  pound-inches.  From  Art.  7  the  allowable 
value  of  S  for  a  variable  load  is  about  10  ooo  pounds  per  square 

inch;  then, 

I/c=  24  000/10  000=  2.4  inches3 

An  infinite  number  of  cross-sections  may  be  selected  having 
this  value  of  I/c.  If  the  section  is  round  and  of  diameter  d, 
it  is  known  that  c  =  \d  and  I  =  -fa7cd*,  hence  ^V7r^3  =  24j  from  which 
^  =  2.9  inches.  If  the  section  is  rectangular,  2  inches  wide  and 
2j  inches  deep,  I/c  is  2.5,  which  is  a  little  too  large,  but  it  would 
be  well  to  use  this  size  because  the  weight  of  the  beam  itself 
has  not  been  considered  in  the  discussion.  The  dimensions 
finally  selected  may  be  investigated  by  Art.  49  in  order  to  deter- 
mine how  closely  the  actual  unit-stress  agrees  with  the  value 
assumed.  Thus,  the  rectangular  section  2X2^  inches  weighs 
17  pounds  per  foot;  the  maximum  moment  is  then  30  530  pound- 
inches  and  the  unit-stress  is  found  to  be  1 1  800  pounds  per  square 
inch,  which  is  18  percent  larger  than  the  allowable  value;  a  larger 
size  than  2X2^  inches  is  hence  required,  and  2X3  inches  will 
be  found  to  be  larger  than  necessary. 

When  the  design  of  a  structure  involves  rolled  I  beams,  the 
computation  of  the  maximum  value  of  M/S  is  made  as  before, 


ART.  51  ECONOMIC  SECTIONS  127 

and  the  corresponding  number  is  sought  in  that  column  of 
Table  6  which  is  headed  I/c.  For  example,  take  a  floor  which 
is  required  to  carry  a  uniform  load  of  180  pounds  per  square 
foot  including  its  own  weight,  this  weight  being  brought  upon 
rolled  steel  beams  which  are  of  24  feet  span  and  spaced  4  feet 
apart.  It  is  required  to  find  what  size  of  beam  should  be  used, 
the  allowable  unit-stress  S  being  specified  as  16  ooo  pounds  per 
square  inch.  First,  the  total  uniform  load  on  one  beam  is  found 
to  be  1^  =  180X24X4  =  17280  pounds;  second,  the  maximum 
moment  is  M  =  JX  17  280X24X12  inch-pounds;  third,  from 
these  If 75  =  38.9,  which  is  the  required  value  of  I/c\  fourth, 
Table  6  shows  that  the  12 -inch  beam  weighing  40  pounds  per 
foot  has  7/^  =  44.8,  and  this  is  the  size  to  be  selected.  The  larger 
table  given  in  the  handbook  of  the  manufacturers  indicates, 
however,  that  a  12 -inch  beam  weighing  35  pounds  per  foot  may 
be  obtained  by  special  order  and  that  its  value  of  I/c  is  38.0; 
whether  this  would  be  cheaper  than  the  1 2-inch  beam  weighing 
40  pounds  per  foot  can  be  determined  only  by  asking  quotations 
of  prices. 

Prob.  500.  Aioo-lb.  1 5-inch  steel  I  beam  of  12  feet  span  sustains 
a  uniformly  distributed  load  of  41  net  tons.  Find  its  factor  of  safety. 
Also  the  factor  of  safety  for  a  24-foot  span  under  the  same  load. 

Prob.  5Qb.  A  floor,  which  is  to  sustain  a  uniform  load  of  175  pounds 
per  square  foot,  is  to  be  supported  by  heavy  lo-inch  steel  I  beams  of 
15  feet  span.  Find  their  proper  distance  apart  from  center  to  center, 
so  that  the  maximum  fiber  stress  may  be  12  ooo  pounds  per  square 
inch. 

ART.  51.    ECONOMIC  SECTIONS 

The  two  fundamental  objects  to  be  secured  in  designing 
engineering  structures  are  stability  and  economy  (Art.  7).  In 
the  case  of  a  beam,  proper  security  will  be  attained  when  the 
horizontal  unit-stress  S  does  not  exceed  that  allowable  in  good 
practice,  and  economy  will  be  secured  by  giving  such  propor- 
tions to  the  cross-section  that  S  is  not  less  than  the  allowable 
value.  Both  stability  and  economy  will  hence  usually  be  pro- 
moted by  making  the  beam  of  such  a  size  that  the  horizontal 


128  CANTILEVER  AND  SIMPLE  BEAMS  CHAP.  VI 

unit-stress  S  has  the  given  allowable  value  at  the  dangerous 
section  when  the  beam  is  fully  loaded.  There  is,  however, 
another  important  idea  to  be  considered,  namely,  the  shape  of 
the  section  should  be  such  that  the  weight  of  the  beam  shall  be 
as  small  as  possible,  and  this  will  be  attained  by  making  the 
section  area  a  as  small  as  possible  and  yet  keep  the  unit-stress  S 
at  the  given  allowable  value. 

Since  the  horizontal  unit-stresses  in  the  section  increase  from 
the  neutral  surface  to  the  upper  and  lower  sides  of  the  beam, 
it  is  evident  that  a  deep  section  will  in  general  require  less  area 
to  furnish  a  given  unit-stress  S  than  a  shallow  one.  Thus,  for  a 
rectangular  section  of  breadth  b  and  depth  d,  the  section  factor 
I/c  is  i^bd3/^d  or  Jfo/2,  so  that  the  flexure  formula  becomes 
%bd2  =  M/S  or  a  =  6M/Sd,  so  that  for  given  values  of  M  and  Sy 
the  section  area  a  will  be  rendered  small  by  making  the  depth  d 
large.  The  depth,  however,  should  not  be  made  so  great  as 
to  give  a  thin  section,  for  this  would  be  deficient  in  lateral  stiffness. 
The  proper  ratio  between  b  and  d  is  governed  by  engineering  prec- 
edent and  practice;  an  extreme  limit  is  perhaps  that  found  in 
wooden  floor  joists  where  the  depth  is  about  six  times  the  breadth. 

Iron  and  steel  beams  may  be  cast  or  rolled  with  almost  any 
desired  shape  of  section.  Cast-iron  T  and  I  beams  came  into 
use  early  in  the  nineteenth  century;  after  1840  wrought-iron 
rolled  beams  gradually  replaced  those  of  cast-iron  for  railroad 
use,  and  these  in  turn  gave  way  after  1890  to  rolled  steel  beams. 
In  these  forms  the  sections  have  such  a  shape  that  the  section 
area  is  the  least  possible  for  a  given  required  unit-stress  S,  and 
this  is  accomplished  by  concentrating  most  of  the  material  in 
the  flanges  and  thus  having  the  larger  part  of  the  section  area 
as  far  from  the  neutral  axis  as  practicable.  The  flexure  formula 
SI/c  =  M  shows  that  this  practice  is  correct,  for,  under  a  given 
bending  moment  Mt  the  unit-stress  S  will  be  made  small  by 
making  I/c  as  large  as  possible,  and  from  the  definition  of  the 
moment  of  inertia  (Art.  43)  it  is  clear  that  /  will  be  rendered 
large  by  placing  the  material  as  far  as  practicable  from  the  neu- 


ART.  51  ECONOMIC  SECTIONS  129 

tral  axis.  The  value  of  /  may  be  expressed  by  ar2,  where  a  is 
the  section  area  and  r  is  the  radius  of  gyration  of  the  section, 
that  is,  the  distance  from  the  neutral  axis  to  the  point  where 
all  the  section  area  might  be  concentrated  and  have  the  same 
moment  of  inertia  as  the  actual  distributed  area.  Inserting 
this  for  /  in  the  flexure  formula,  it  becomes  ar2/c  =  M/S  and 
thus  it  is  plain  that  a  may  be  rendered  small  by  making  r2/c 
as  large  as  possible.  Since  r  is  always  less  than  c,  the  placing 
of  the  greater  part  of  the  section  in  the  flanges,  while  the  web 
is  made  thin,  renders  r2/c  much  larger  than  for  a  rectangular 
section,  and  thus  a  is  made  smaller  and  economy  of  material  is 
secured. 

For  example,  take  the  smallest  1 5-inch  I  beam  in  Table  6 
for  which  a=  12.48  square  inches,  1  =  441.7  inches4  and  I/c  = 
58.9  inches3.  A  square  section  of  the  same  strength  must  have 
the  same  value  of  I/c,  whence  ^d3  =  $8.C)  and  0  =  50.0  square 
inches,  which  is  four  times  the  section  area  of  the  I  and  hence 
the  square  beam  will  weigh  four  times  as  much  as  the  I  beam, 
and  its  cost  will  be  four  times  as  much  if  the  price  per  pound 
is  the  same. 

For  wrought  iron  and  structural  steel  the  ultimate  strengths 
in  tension  and  compression  and  the  allowable  unit-stresses 
are  usually  the  same;  hence  the  flanges  of  beams  of  these 
materials  are  made  equal  in  size.  Cast-iron,  however,  has  a 
much  higher  ultimate  strength  in  compression  than  in  tension 
and  hence  the  tensile  flange  should  have  the  larger  area.  The 
proper  relative  proportions  of  the  flange  areas  of  cast-iron  beams 
have  never  been  definitely  established,  and  such  beams  are 
now  rarely  used  except  in  unimportant  buildings. 

The  strongest  rectangular  beam  that  can  be  cut  from  a  cir- 
cular log  will  be  that  which  has  the  largest  section  factor  I/c. 
If  b  be  the  breadth  and  d  the  depth,  the  section  factor  is  %bd2, 
and  bd2  is  to  be  made  a  maximum;  or,  if  D  be  the  diameter  of 
the  log,  b(D2  —  b2)  is  to  be  made  a  maximum.  Differentiating 
this  and  equating  the  derivative  to  zero,  gives, 

whence  d=D\/% 


130  CANTILEVER  AND  SIMPLE  BEAMS  CHAP.  VI 

Hence  b  is  to  d  as  5  to  7  nearly.     From  this  it  is   easy  to  show 

that  the  way  to  lay  off  the  strongest  rectangular  beam  on  the 
end  of  a  circular  log  is  to  divide  the  diameter 
into  three  equal  parts,  from  the  points  of  divi- 
sion draw  perpendiculars  to  the  circumference, 
and  then  join  the  points  of  intersection  with 
the  ends  of  the  diameter,  as  shown  in  the  figure. 
The  rectangular  beam  thus  cut  out  is,  of  course, 
Flg<  51  not  as  strong  as  the  log,  and  the  ratio  of  its 

strength,  to  that  of  the  log  is  that  of  their  values  of  I/c,  which 

will  be  found  to  be  about  0.65. 

Prob.  51.  Cast-iron  beams  with  a  «-j  section  are  sometimes  used  in 
buildings.  Let  the  thickness  be  uniformly  one  inch,  the  base  8  inches, 
the  height  6  inches,  and  the  span  12  feet.  Find  the  unit-stresses  5 
and  S'  at  the  dangerous  section  under  a  uniform  load  of  5  ooo  pounds. 

ART.  52.    RUPTURE  OF  BEAMS 

The  flexure  formula  S.I/c  =  M  is  only  true  for  stresses 
within  the  elastic  limit  of  the  material,  since  beyond  that  limit 
the  latter  part  of  law  6  of  Art.  40  does  not  hold.  Experience 
shows  that  the  elongations  and  shortenings  of  the  horizontal 
fibers  are  proportional  to  their  distances  from  the  neutral  axis 
when  the  stresses  exceed  the  elastic  limit,  but  these  changes  of 
length  increase  in  a  more  rapid  ratio  than  the  unit-stresses  (Arts. 
4-5).  Hence  the  unit-stresses  of  tension  and  compression  in- 
crease in  a  less  rapid  ratio  than  their  distance  from  the  neutral 
axis  for  all  fibers  where  the  elastic  limit  is  exceeded.  Thus 
Fig.  52a  represents  a  rectangular  cast-iron  beam  where  the 
tensile  unit-stresses  below  n  have  exceeded  the  elastic  limit, 
while  Fig.  526  shows  a  T  section  under  similar  conditions. 
Here  the  algebraic  sum  of  all  the  horizontal  stresses  is  zero,  but 
the  neutral  surface  does  not  pass  through  the  center  of  gravity 
of  the  section,  because  the  unit-stresses  are  not  proportional  to 
their  distances  from  that  surface  as  is  required  in  the  demon- 
stration of  Art.  40. 

When  a  beam  is  loaded  so  heavily  that  any  fiber  on  one  or 


RUPTURE  OF  BEAMS 


131 


both  sides  of  the  neutral  surface  are  stressed  above  the  elastic 
limit,  the  flexure  formula  S.I/c  =  M  is  not  valid,  and  a  value 
of  S  computed  from  it  is  not  correct.  It  is,  however,  very  cus- 
tomary to  use  this  formula  for  the  rupture  of  beams,  but  in  so 
doing  it  must  not  be  forgotten  that  it  is  so  used  merely  for  con- 
venience of  making  comparisons  and  not  on  account  of  any 
theoretical  basis. 


Fig.  52a 


Fig.  526 


When  a  beam  is  ruptured  under  transverse  loads  the  value 
of  S  computed  from  the  flexure  formula  is  called  the  '  modulus 
of  rupture '  of  the  material.  If  this  formula  were  valid 
beyond  the  elastic  limit,  the  value  of  S  for  rupture  would  agree 
with  the  least  ultimate  strength  of  the  material,  with  tension 
in  the  case  of  cast  iron  and  with  compression  in  the  case  of  timber. 
It  is,  however,  always  found  that  this  computed  value  does  not 
agree  with  either  the  ultimate  tensile  or  compressive  strength 
of  the  material  but  is  intermediate  between  them.  This  quantity 
is  not  a  physical  constant,  but  a  figurative  value  computed  from 
an  incorrect  formula,  and  hence  is  mainly  valuable  for  rough 
comparative  purposes.  It  will  be  designated  by  Sf,  and  the 
following  are  its  approximate  average  values  in  pounds  per 
square  inch  as  determined  by  experiment: 


Tensile  Modulus  of     Compressive 

Strength,  St       Rupture,  S}      Strength,  Se 

3  000 
6  ooo 
8000 
90000 


Material 
Brick 
Stone 

Timber  10  ooo 

Cast  Iron  20  ooo 

Wrought  Iron  50  ooo 

Structural  Steel  60  ooo 

Strong  Steel  100  ooo 

For  wrought  iron  and  structural  steel  no  values  of  Sf  are  given 


800 

2  OOO 

9  ooo 

35000 


no  ooo 


50  ooo 
60  ooo 

120  OOO 


132  CANTILEVER  AND  SIMPLE  BEAMS  CHAP,  vi 

because  beams  of  these  materials  bend  indefinitely  under  increas- 
ing transverse  loads,  so  that  failure  does  not  occur  by  breaking. 
In  fact,  bars  of  wrought  iron  and  soft  steel  may  be  bent  double 
without  breaking  (Arts.  24  and  25). 

By  the  use  of  the  above  experimental  values  of  Sf  it  is  easy, 
with  the  help  of  the  formula  Sf .  I/c  =  M,  to  determine  what 
load  will  cause  the  rupture  of  a  given  beam,  or  what  must  be 
its  length  or  size  in  order  that  it  may  rupture  under  assigned 
loads.  The  formula  when  used  in  this  manner  is  entirely  em- 
pirical and  has  no  rational  basis.  As  an  example,  let  it  be  required 
to  find  the  length  of  a  cast-iron  cantilever  beam,  2X2  inches  in 
section,  in  order  that  it  may  break  at  the  fixed  end  under  its 
own  weight.  The  weight  of  the  beam  is  7^  =  2X2X3^X0.94  = 
12.53  pounds  per  linear  foot  or  1.044  pounds  per  linear  inch; 
the  bending  moment  is  M  =  JX i. 044/2  =  0.522^  pound-inches, 
where  /  is  in  inches;  the  value  of  I/c  is  %bd2  =  1.333  inches3. 
Inserting  these  in  the  formula  it  becomes  0.522/2  =  35  000X1.333, 
from  which  7  =  299  inches  or  about  25  feet. 

The  modulus  of  rupture  Sf  is  sometimes  called  '  computed 
flexural  strength.'  The  latter  points  out  the  meaning  of  the 
quantity,  but  the  former  is  in  general  use. 

Prob.  520.  Compute  the  size  of  a  square  wooden  simple  beam  of 
8  feet  span  in  order  to  break  under  its  own  weight. 

Prob.  526.-  A  cast-iron  simple  beam  2  inches  square  and  12  feet 
long  carries  two  equal  loads  at  the  quarter  points.  Find  the  loads 
which  will  cause  rupture. 

ART.  53.     MOVING   LOADS 

The  loads  upon  a  beam  consist  of  its  own  weight  and  the 
weight  of  the  uniform  or  concentrated  loads  which  it  carries. 
These  are  called  '  Dead  Loads '  when  they  are  permanent  in 
position  and  'Live  Loads'  when  they  may  move.  Beams  used 
in  buildings  are  subject  to  the  dead  load  of  the  floor  and  to  the 
live  load  of  a  crowd  of  people;  beams  used  in  bridges  are  sub- 
ject to  the  dead  load  of  their  own  weight  and  to  other  permanent 


ART.  53 


MOVING  LOADS 


133 


loads,  but  they  receive  the  greatest  stress  from  the  live  load  of 
moving  wheels  or  of  crowds  of  people.  In  the  preceding  articles 
dead  loads  have  alone  been  generally  considered,  but  it  has  been 
recognized  that  the  maximum  moment  due  to  a  single  concen- 
trated load  on  a  simple  beam  occurs  when  it  is  placed  at  the 
middle  of  the  span.  Other  cases  of  concentrated  loads  will 
now  be  discussed. 


f  r 


Fig.  53a 


Fig.  536 


When  two  wheels  which  are  at  fixed  distance  apart,  like  two 
wagon-wheels  on  separate  axles,  roll  over  a  beam,  it  might  be 
thought  that  the  greatest  moment  due  to  them  would  occur  when 
they  are  on  opposite  sides  of  the  middle  and  equally  distant 
from  it,  as  in  Fig.  53#,  but  this  is  not  the  case.  To  find  the 
position  which  will  give  the  greatest  moment,  let  /  be  the  span, 
p  the  distance  between  the  loads,  and  z  the  distance  from  the 
left  support  to  one  of  the  loads,  as  in  Fig.  536.  The  maximum 
moment  will  occur  under  one  of  these  loads,  as  the  shear  and 
moment  diagrams  show.  The  left  reaction  is  found  from 
Ril=P(l  —  p  —  z)+P(l—z)  and  the  moment  under  the  first  load 
then  is  M  =  (P/l)(2lz  —  pz  —  2Z2).  By  the  usual  method,  the 
value  of  z  which  renders  M  a  maximum  is  found  to  be  z  =  J/— \p, 
so  that  the  center  of  gravity  of  the  two  loads  is  as  far  to  the  right 
of  the  middle  as  the  dangerous  section  is  to  the  left  of  the  middle. 
For  example,  let  each  load  be  3  coo  pounds,  their  distance  apart 
be  5  feet,  and  the  span  be  15  feet;  then  2  =  6.25  feet,  the  left  reac- 
tion is  Ri  =  2  500  pounds,  and  the  maximum  bending  moment 
is  M  =  2  500X6.25  =  15  625  pound-feet.  If  these  loads  are 
placed  as  in  Fig.  53a,  the  reaction  is  3  ooo  pounds  and  the 
moment  is  3000X5.0  =  15000  pound-feet. 


134 


CANTILEVER  AND  SIMPLE  BEAMS 


CHAP.  VI 


When  two  unequal  loads  PI  and  P^  roll  over  a  simple  beam, 
let  z  be  the  distance  from  the  left  support  to  PI,  and  p  be  the 
distance  from  the  greater  load  PI  to  the  smaller  one  P%.  Then, 
proceeding  as  before,  it  is  easy  to  show  that  the  maximum  moment 
occurs  under  the  first  load  when  the  distance  between  the  first 
load  and  the  center  of  gravity  of  the  two  loads  is  bisected  by 
the  middle  of  the  span.  When  there  are  three  or  more  con- 
centrated loads,  the  section  of  maximum  moment  does  not  always 
lie  under  the  first  load,  but  the  general  rule  always  holds  good 
that  the  distance  between  that  section  and  the  center  of  gravity 
of  the  loads  is  bisected  by  the  middle  of  the  span. 


Fig.  53c 


Fig.  53d 


When  a  uniform  live  load  moves  over  a  simple  beam  there 
is,  for  any  given  position  of  the  same,  a  section  of  maximum 
moment,  this  being  where  the  shear  passes  through  zero  (Art.  47). 
Such  cases,  with  their  shear  and  moment  diagrams,  are  shown 
in  the  two  figures  above,  but  it  is  rarely  necessary  to  compute 
these  moments,  because  the  absolute  maximum  moment  occurs 
when  the  uniform  live  load  covers  the  entire  beam,  and  this  is 
liable  to  take  place  at  any  time.  It  is,  however,  of  great  impor- 
tance that  the  student  should  be  able  readily  to  draw  the  shear 
and  moment  diagrams  for  any  assigned  position  of  the  live  load, 
whether  it  be  uniform  or  partly  uniform  and  partly  concentrated. 

The  maximum  shear  due  to  a  live  load  occurs  when  the  load 
is  placed  so  as  to  give  the  greatest  shear  at  one  of  the  supports. 
Thus  the  maximum  shear  due  to  the  live  load  in  Fig.  536  occurs 
at  one  of  the  supports  when  one  load  is  just  about  to  pass  upon 
that  support  and  its  numerical  value  is  2P—P.  p/l. 

By  the  use  of  the  shear  formula  5>  =  V  and  the  flexure  formula 
SI/c=M,  the  unit-stresses  Sg  and  S  are  computed  for  live  loads, 


ART.  54  DEFLECTION   OF   CANTILEVER   BEAMS  135 

after  the  maximum  values  of  V  and  M  have  been  found,  in  the 
same  manner  as  if  the  loads  were  dead.  Live  load,  however, 
really  produces  greater  stresses  than  dead  load  and  hence  the 
computed  values  of  S8  and  S  are  increased  in  practice  accord- 
ing to  the  rules  stated  in  Art.  136. 

Prob.  53a.  Three  loads,  spaced  4  feet  apart,  one  being  3  ooo  and 
the  others  i  500  pounds,  roll  over  a  simple  beam  of  21  feet  span.  Find 
the  position  of  these  loads  which  will  give  the  maximum  moment  and 
compute  its  value. 

Prob.  536.  For  Fig.  536  find  the  position  of  the  live  load  which 
gives  the  maximum  shear  at  the  middle  of  the  span  and  compute  its 
value.  Find  also  the  position  which  gives  maximum  shear  at  the 
quarter  point  of  the  span  and  compute  its  value. 


ART.  54.     DEFLECTION  OF  CANTILEVER  BEAMS 

In  Art.  45  the  differential  equation  of  the  elastic  curve  was 
deduced  and  the  general  method  of  applying  it  to  a  particular 
case  was  indicated.  The  origin  of  coordinates  may  be  taken 
at  any  point  in  the  plane,  but  for  a  cantilever  beam  it  is  most 
convenient  to  take  it  at  the  free  end,  since  the  algebraic  work 
is  thus  simplified.  In  the  equation  El  .  d2y/dx2=M,  the  bend- 
ing moment  M  is  to  be  expressed  in  terms  of  the  abscissa  x,  and 
by  two  integrations  the  equation  between  y  and  x  is  deduced. 

Case  I.  Uniform  Load.  —  Let  x  and  y  be  the  coordinates 
of  the  elastic  curve  with  respect  to  rectangular  axes  through 
the  free  end  of  the  beam,  as  in  Fig.  54a.  Let  the  uniform 
load  per  linear  unit  be  w\  then  for  any  section  M  =  —  %wx2,  and 
the  general  formula  becomes  El  .  d2y/dx2  =  —  \wx2.  Integrating 
this,  there  is  found, 


in  which  C  is  the  constant  of  integration  and  dy/dx  is  the  tan- 
gent of  the  angle  which  the  tangent  to  the  elastic  curve  makes 
with  the  axis  of  abscissas.  To  determine  this  constant,  con- 
sider that  dy/dx  becomes  zero  when  x  equals  the  length  of  the 


136  CANTILEVER  AND  SIMPLE  BEAMS  CHAP.  VI 

beam,  /;  hence  the  value  of  C  is  %wl3  and  then, 


Integrating  again,   and   determining  the   constant   by   the   con- 
dition that  y  equals  zero  when  x  equals  zero,  there  results, 


which  is  the  equation  of  the  elastic  curve.  When  x=l,  the 
value  of  y  is  the  deflection  of  the  end  of  the  beam  below  that 
at  the  wall  and  this  will  be  designated  by/.  Accordingly, 

f=%wl4/EI  or  f=$WP/EI 

is  the  deflection  of  the  end  of  a  cantilever  beam  under  the  uniform 
load  W,  if  the  elastic  limit  of  the  material  be  not  exceeded. 


Fig.  54<z  Fig.  54& 

Case  II.  Load  P  at  the  Free  End. — Take  the  origin  at  the 
free  end  as  in  Fig.  54&,  and  let  x  and  y  be  the  coordinates  of 
the  elastic  curve  at  any  section.  The  bending  moment  M  is 
-Px,  and  the  general  equation  is  El .  d2y/dx2  =  -  Px.  By 
integration  there  results  El .  dy/dx  =  -%Px2  +  C  and  C  is  deter- 
mined by  the  condition  that  the  tangent  dy/dx  becomes  zero 
when  x  =  l.  Hence  El .  dy/dx  =  ±P(l2-x2),  and  the  second 
integration  gives  for  the  elastic  curve  El .  y  =  %Pl2x  —  $Px3,  the 
constant  being  zero  because  y  becomes  zero  when  x  is  zero. 
When  x  =  l  the  value  of  y  is  the  maximum  deflection/,  and/= 
^P/3/£7,  which  is  2§  times  as  great  as  the  deflection  due  to 
the  same  load  uniformly  distributed  over  the  length. 

Case  III.  Load  P  at  any  Point. — Let  d  be  the  distance  of 
the  load  from  the  left  end  as  in  Fig.  54c,  where  /  is  the  length 
of  the  beam  and  K  any  number  less  than  unity.  Take  the  origin 
of  coordinates  under  the  load,  then  by  the  preceding  case  the 
deflection  under  the  load  is  JP(/-/c/)3/£/.  The  free  end,  how- 
ever, is  lower  than  the  load  and  since  M  =o  on  the  left  of  the 
load,  the  radius  of  curvature  of  the  elastic  curve  is  there  infinite 


ART.  54  DEFLECTION    OF   CANTILEVER   BEAMS  137 

(Art.  45)  and  that  curve  is  a  straight  line.  Let  tan  6  be  the  tan- 
gent of  the  angle  which  the  tangent  to  the  elastic  curve  under 
the  load  makes  with  the  horizontal;  then  the  free  end  is  lower 
than  the  load  by  the  amount  id  tan#.  The  value  of  tan#  is  deter- 
mined from  Case  II,  by  making  x  =  o  and  1  =  1  -id  in  the  expres- 
sion for  dy/dx,  and  thus  the  value  of  d  tanfl  is  found  to  be 
JP/3/c(i  —K)2/EI.  Therefore,  by  adding  the  two  quantities, 


which  is  the  deflection  of  the  free  end  due  to  the  given  load. 
When  ic  =  i,  the  load  is  at  the  wall  and  /=o;  when  K=O  the 
load  is  at  the  free  end  and  /  becomes  the  same  as  in  the  pre- 
ceding case. 


Fig.  54c 

Case  IV.  Several  Loads.  —  For  a  uniform  load  and  a  load  P 
at  the  end,  the  value  of  M  is  —  %wx2—  Px.  By  integration  it  is 
found  that  the  ordinate  y  is  the  sum  of  those  due  to  W  and  P 
separately,  and, 


is  the  deflection  of  the  free  end.  Similarly,  for  any  number  of 
loads,  the  deflection  is  the  sum  of  the  deflections  due  to  the  loads 
taken  separately;  hence,  as  in  cases  of  axial  stress,  each  load 
produces  its  effect  independently  of  others.  In  order  that  this 
deflection  may  be  found  correctly  from  the  formulas,  it  is 
necessary  that  the  maximum  unit-stress  S  computed  for  all  the 
loads  from  the  flexure  formula  S  .I/c=M  shall  not  exceed  the 
elastic  limit  of  the  material. 

In  all  cases  the  deflection  of  a  cantilever  beam  of  uniform 
section  varies  directly  as  the  load  and  the  cube  of  the  length, 
and  inversely  as  E  and  /.  For  a  rectangular  section,  I=-f^bd3, 
so  that  the  deflection  varies  inversely  as  the  breadth  and  inversely 
as  the  cube  of  the  depth.  These  laws  also  hold  for  the  simple 
beams  discussed  in  the  next  article. 


13S  CANTILEVER  AND  SIMPLE  BEAMS  CHAP.  VI 

Prob.  540.  In  order  to  find  the  modulus  of  elasticity  of  a  cast-iron 
bar  2  inches  wide,  4  inches  deep,  and  6  feet  long,  it  was  balanced  upon 
a  support  and  a  weight  of  4  ooo  pounds  hung  at  each  end,  causing  a 
deflection  of  0.401  inches.  Compute  the  value  of  E. 

Prob.  54ft.  A  wooden  cantilever,  3  inches  wide,  4  inches  deep,  and 
15  feet  long,  carries  two  equal  loads  as  shown  in  Fig.  54d,  one  being 
5  feet  from  the  end  and  the  other  10  feet  from  the  end.  Compute  the 
weight  of  these  loads  so  that  the  maximum  unit-stress  S  may  be  two- 
thirds  of  the  elastic  limit.  Compute  the  deflection  at  the  end  due  to 
the  two  loads. 

ART.  55.     DEFLECTION  OF  SIMPLE  BEAMS 

The  deflection  of  a  simple  beam  due  to  a  load  at  the  middle, 
or  to  a  uniform  load,  is  readily  obtained  from  the  expressions 
just  deduced  for  cantilever  beams.  Thus,  for  a  simple  beam 
of  span  /  with  a  load  P  at  the  middle,  let  Fig.  55a  be  inverted 
and  it  will  be  seen  to  be  equivalent  to  two  cantilever  beams  of 
length  \l  with  a  load  \P  at  each  end.  The  formula  for  the  maxi- 
mum deflection  of  a  cantilever  beam  hence  applies  to  this  figure, 
if  /  be  replaced  by  \l  and  P  by  JP  which  gives  /  =  PP/4SEI  for 
the  deflection  at  the  middle  of  the  simple  beam.  It  will  be  well, 
however,  to  use  the  general  formula  (45)  and  discuss  each  case 
independently. 

Case  I.  Uniform  Load.  —  Let  w  be  the  load  per  linear  unit 
and  x  the  distance  of  any  section  from  the  left  end.  For  this 
section  M  =  \wlx  —  \wx2  and  the  differential  equation  of  the 
elastic  curve  is, 


Integrate  this  and  find  the  constant  by  the  condition  that  the 
tangent  dy/dx  equals  zero  when  x  is  J/;    then, 


ox 

Integrating  again  and  determining  the  constant  by  the  condition 
that  y  is  zero  when  x  is  zero,  there  is  found, 


ART.  55  DEFLECTION   OF    SIMPLE    BEAMS  139 

Now,  making  x  =  £/,  the  value  of  y  is  the  deflection  /,  which  is 
negative  because  it  is  measured  downward  from  the  axis  of 
abscissas  through  the  supports.  It  is,  however,  unnecessary 
to  write  this  sign,  and  hence, 

/-ih  ™P/EI  or  f=lslIWP/EI 

is  the  elastic  deflection  of  the  simple  beam  under  the  uniform 
load  W. 

Case  II.  Load  P  at  the  Middle  —  As  before  let  the  origin 
be  taken  at  the  left  support,  as  in  Fig.  55a.  For  any  section  between 
the  left  support  and  the  middle  M  =  %Px  and  the  differential 
equation  of  the  elastic  curve  is  El  .  d2y/dx2  =  %Px.  Integrate 
this  and  find  the  constant  by  the  evident  condition  that 
dy/dx=o  when  x  =  %l.  Then  integrate  again  and  find  the  con- 
stant by  the  fact  that  y=o  when  x=o.  Thus, 


is  the  equation  of   elastic  curve  between  the  left-hand  support 
and  the  load.      For  the  greatest  deflection  make  #  =  £/,  then, 


is  the  deflection  due  to  the  single  load  P  at  the  middle,  which 
is  1.6  times  as  great  as  that  due  to  the  same  uniform  load. 

IP 

---  Kl  ------- 


Fig.  55a  Fig.  55b 

Case  III.  Load  P  at  any  Point. — Here  the  expressions  for 
the  moment  M  are  different  on  opposite  sides  of  the  load,  and 
hence  there  are  two  elastic  curves  which  have  distinct  equations 
but  which  have  a  common  tangent  and  ordinate  under  the  load. 
As  in  Fig.  55&  let  the  load  be  placed  at  a  distance  /c/  from  the 
left  support,  K  being  a  number  less  than  unity.  Then  the  left 
reaction  is  R=P(i—ic).  From  the  general  equation  (45),  with 
the  origin  at  the  left  support,  there  are  found, 
On  the  left  of  the  load, 

(a)        £*§ 

(c) 


140  CANTILEVER  AND  SIMPLE  BEAMS  CHAP,  vi 

On  the  right  of  the  load, 
(ay        EI~=Rx-P(x-Kl) 


(by          £/ 

(cy  EIy=  \R^-\Po^+  $PKlx2+  C3x+  C4 

To  determine  the  constants  consider  in  (c)  that  y  =  o  when  x  =  o, 
and  hence  that  C2=o.  Also  in  (cy,  y=o  when  x  =  l\  again, 
since  the  curves  have  a  common  tangent  under  the  load,  (6) 
equals  (b)f  when  x  =  Kl,  and  since  they  have  a  common  ordinate 
at  that  point  (c)  equals  (c)f  when  x  =  K.l.  Or, 


From  these  three  conditions  the  values  of  C\,  €3,  and  C4  are 
determined.  Then  the  equation  of  the  elastic  curve  on  the  left  of 
the  load  is  found  to  be, 


To  ascertain  the  maximum  deflection,  the  value  of  x  which  ren- 
ders y  a  maximum  is  to  be  obtained  by  equating  the  first  deriva- 
tive to  zero.  If  K  is  greater  than  J,  this  value  of  x  inserted  in  the 
above  equation  gives  the  maximum  deflection;  if  K  is  less  than 
J,  the  maximum  deflection  is  on  the  other  side  of  the  load.  For 
instance,  if  *  =  f,  the  equation  of  the  elastic  curve  on  the  left 
of  the  load  is,  ^^Ely^^Po^—  i$Pl2x,  and  y  has  its  maximum 
value  when  x=o.$$()l.  The  greatest  possible  deflection  due  to 
a  single  load  occurs  when  it  is  at  the  middle  of  the  span  and  its 
value  is  that  deduced  in  Case  II. 

Case  IV.  Several  Loads.  —  Here,  as  for  cantilevers,  the 
deflection  due  to  several  loads  is  obtained  by  taking  the  sum  of 
the  several  deflections;  but  it  must  be  carefully  noted  that  the 
computed  value  will  be  correct  only  when  the  unit-stress  S  at 
the  dangerous  section  is  less  than  the  elastic  limit  of  the  material. 
The  above  formula  for  the  deflection  due  to  a  load  at  the  middle 
is  frequently  used  to  determine  the  modulus  of  elasticity  E,  several 


ART.  56  COMPARATIVE   STRENGTH   AND    STIFFNESS  141 

values  of/  being  measured  for  several  different  loads,  in  order  to 
obtain  a  mean  value  of  E. 

Prob.  550.  When  K  is  greater  than  J  in  Fig.  556,  show  that  the 
maximum  deflection  is/=P/3(i-/c)(§K-J/c2)V3£/. 

Prob.  556.  In  order  to  find  the  modulus  of  elasticity  of  Quercus 
alba,  a  bar  4  centimeters  square  and  one  meter  long  was  supported  at 
the  ends,  and  loaded  in  the  middle  with  weights  of  50  and  100  kilo- 
grams, measured  deflections  being  6.6  and  13.0  millimeters.  Com- 
pute the  mean  value  of  E  in  kilograms  per  square  centimeter. 

ART.  56.     COMPARATIVE  STRENGTH  AND  STIFFNESS 

The  strength  of  a  bar  under  tension  is  measured  by  the  load 
that  it  can  carry  with  an  assigned  unit-stress.  In  the  same 
manner  the  strength  of  a  beam  is  measured  by  the  load  that 
it  can  carry  with  an  assigned  unit-stress  on  the  remotest  fiber  at  the 
dangerous  section.  Let  it  be  required  to  determine  the  relative 
strength  of  the  four  following  cases, 

ist,  A  cantilever  loaded  at  the  end  with  W 

2nd,  A  cantilever  uniformly  loaded  with  W 

3rd,  A  simple  beam  loaded  at  the  middle  with  W 

4th,  A  simple  beam  loaded  uniformly  with  W 

Let  /  be  the  length  in  each  case.  Then,  from  Art.  47  and  the 
flexure  formula  M  =  S  .  I/c,  there  is  found, 

fonst,  M=Wl  and  hence  W=SI/d 

for  2nd,  M=\Wl  and  hence  W=2  .  Sl/d 

for3rd,  M=\Wl  and  hence  W=4-SI/d 

for  4th,  M=\Wl  and  hence  W=8  .  Sl/d 

Therefore  the  comparative  strengths  of  the  four  cases  are  as 
the  numbers  i,  2,  4,  8.  That  is,  if  four  such  beams  be  of  equal 
size  and  length  and  of  the  same  material,  the  second  is  twice  as 
strong  as  the  first,  the  third  four  times  as  strong,  and  the  fourth 
eight  times  as  strong.  From  these  equations  also  result  the 
following  important  laws: 

The  strength  of  a  beam  varies  directly  as  S,  directly  as  /, 
inversely  as  c,  and  inversely  as  the  length  /. 


142  CANTILEVER  AND  SIMPLE  BEAMS  CHAP.  VI 

A  load  uniformly  distributed  produces  only  one-half  as 
much  stress  as  the  same  load  when  concentrated. 

These  apply  to  all  cantilever  and  simple  beams  whatever  be 
the  shape  of  the  cross-section. 

When  the  cross-section  is  rectangular,  let  b  be  the  breadth 
and  d  the  depth,  then  the  value  of  I/c  is  %bd2  and  the  above  equa- 
tions become  W  =  aSbd2/6l,  where  the  number  a  is  i,  2,  4,  or  8 
as  the  case  may  be.  Therefore, 

The  strength  of  a  rectangular  beam  varies  directly  as  its 
breadth,  directly  as  the  square  of  its  depth,  and  inversely  as  its 
length. 

The  reason  why  rectangular  beams  are  put  with  the  greatest 

dimension  vertical  is  thus  again  shown. 

In  the  above  equations  the  load  W  is  the  allowable  load  when 
S  is  the  allowable  unit-stress,  and  W  is  the  load  which  will  rupture 
the  beam  when  5  is  the  fictitious  ultimate  flexural  unit-stress 
whose  mean  values  are  given  in  Art.  52. 

The  stiffness  of  a  bar  under  tension  is  measured  by  the  load 
that  it  can  carry  with  a  given  elongation.  Similarly  the  stiff- 
ness of  a  beam  is  indicated  by  the  load  that  it  can  carry  with  a 
given  deflection.  For  the  two  preceding  articles  the  values  of  W 
for  the  four  common  cases  of  cantilever  and  simple  beams  are» 

for  a  cantilever  loaded  at  the  end,  W=  3  .  Elf /I3 

for  a  cantilever  uniformly  loaded,  W=8  .  Elf /I3 

for  a  simple  beam  loaded  at  middle,  W=4%  .  Elf /I3 

for  a  simple  beam  uniformly  loaded,  PF=-2-f4- .  Elf /I3 

which  show  that  the  relative  stiffness  of  these  four  cases  are  as 
the  numbers  i,  2§,  16,  and  25!. 

These  equations  also  show  that  the  stiffness  of  a  beam,  when 
the  greatest  unit-stress  does  not  exceed  the  elastic  limit  of  the 
material,  varies  directly  as  E  and  /  and  inversely  as  the  cube 
of  the  length.  It  thus  appears  that  the  laws  of  stiffness  are 
very  different  from  those  of  strength.  For  a  rectangular  section 
I  =  ^bd3,  and  hence  the  stiffness  varies  directly  as  the  breadth 
and  directly  as  the  cube  of  the  depth. 


.  57       CANTILEVERS  OF  UNIFORM  STRENGTH  143 

The  four  cases  above  discussed  have  given  the  following 
expressions  for  the  allowable  load;  for  the  strength  of  the  beam 

W=  a  .  SI/cl        where  «=  i,  2,  4,  or  8 
and  for  the  stiffness  of  the  beam, 

W=p  .  Elf  IP      where  /?=3,  8,  48,  or  76£ 

By  equating  these  values  of  W,  the  following  relations  between 
the  unit-stress  S  and  the  deflection  f  are  obtained, 

S/f=pEc/aP  or          f=aSl2/fiEc  (56) 

which  are  only  valid  when  S  is  less  than  the  elastic  limit  of  the 
material  These  equations  show  that  the  deflection  /,  for  similar 
beams  of  the  same  material  under  the  same  unit-stress  S,  varies 
as  l2/c. 

Table  12,  at  the  end  of  this  volume,  recapitulates  the  impor- 
tant facts  regarding  strength,  deflection,  and  stiffness,  which 
have  been  deduced  in  the  preceding  articles. 

Prob.  56a.  Compare  the  strength  of  a  joist  3X8  inches  when  laid 
with  long  side  vertical  with  that  when  it  is  laid  with  short  side  vertical. 
Compare  also  the  stiffnesses. 

Prob.  566.  Find  the  deflection  of  the  lightest  steel  lo-inch  I  beam, 
9  feet  in  span,  when  stressed  by  a  uniform  load  up  to  30  ooo  pounds 
per  square  inch. 

Prob.  56c.  Compare  the  working  strength  of  a  light  p-inch  steel  I 
beam  with  that  of  a  wooden  beam  8X12  inches  in  section,  the  span 
being  the  same  for  both. 

ART.  57.     CANTILEVER  BEAMS  OF  UNIFORM  STRENGTH 

All  beams  thus  far  discussed  have  been  of  constant  cross- 
section  throughout  their  entire  length.  But  in  the  flexure  formula 
S.I/c  =  M,  the  unit-stress  S  is  proportional  to  the  bending 
moment  M,  and  hence  varies  throughout  the  beam  in  the  same 
way  as  the  moments  vary.  Accordingly  some  parts  of  the  beam 
are  but  slightly  stressed  in  comparison  with  the  dangerous  sec- 
tion, and  perhaps  more  material  is  used  than  is  necessary. 

A  beam  of  uniform  strength  is  one  so  shaped  that  the  unit- 
stress  S  is  the  same  in  all  sections  at  the  upper  and  lower  sur- 


144 


CANTILEVER  AND  SIMPLE  BEAMS 


CHAP.  VI 


faces.  Hence  to  ascertain  the  form  of  such  a  beam  the  unit- 
stress  S  must  be  taken  as  constant  and  I/c  be  made  to  vary  with 
M.  The  discussion  will  be  given  only  for  the  simplest  cases, 
namely,  those  where  the  sections  are  rectangular,  the  breadth 
being  b  and  the  depth  d.  For  these  I/c  =  ^bd2,  and  the  flexure 
formula  becomes, 

lSb<P=M  or  bd2=6M/S 

In  this  bd2  must  be  made  to  vary  with  M  in  order  to  give  forms 
of  uniform  strength. 


Side  View 


Plan 

Fig.  57a 


Side  View 


Fig.  57b 


For  a  cantilever  beam  with  a  load  P  at  the  end,  the  value  of  M 
without  regard  to  sign  is  Px  and  the  equation  becomes  %Sbd2  = 
Px,  in  which  P  and  S  are  constant.  If  the  breadth  is  taken  as 
constant,  d2  varies  with  x  and  the  profile  is  that  of  a  parabola 
having  its  vertex  at  the  free  end.  The  depth  di  of  the  beam 
at  the  wall  is  found  from  \Sbd\2  =  Pl,  and  comparing  this  with 
the  first  equation  there  results  the  simpler  form  d  =  diV/oc/l  for 
the  relation  between  d  and  x;  Fig.  57 a  shows  a  profile  plotted 
from  this  equation.  When  the  depth  of  the  cantilever  beam 
is  constant,  then  b  varies  directly  as  x  and  the  plan  of  the  beam 
is  a  triangle,  as  shown  in  Fig.  576;  the  breadth  b\  at  the  wall  is 
found  from  %Sd2bi  =  P/,  and  hence  equation  between  b  and  x 
is  more  simply  expressed  by  b  =  (bi/l)x. 

For  a  cantilever  beam  uniformly  loaded  with  w  per  linear 
unit  M  =  %wx2,  and  the  equation  becomes  %Sbd2  =  %ivx2,  in  which 
w  and  S  are  known.  If  the  breadth  is  taken  as  constant;  then  d 
varies  as  x  and  the  side  view  is  a  triangle,  as  in  Fig.  57c,  where 
the  depth  at  any  point  is  given  by  d=(di/l)x,  the  depth  d\  being 
that  at  the  wall,  which  is  determined  from  ^Sbdi2  =  ^wl2.  If, 
however,  the  depth  is  taken  as  constant,  then  b  varies  as  x2, 


ART.  57 


CANTILEVERS  OF  UNIFORM  STRENGTH 


145 


and  b  may  be  found  from  b  =  bi(x/l)2,  where  bi  is  the  breadth 
at  the  wall;  this  is  the  equation  of  a  parabola  having  its  vertex 
at  the  free  end  and  its  axis  vertical,  or  the  plan  of  the  beam  may 
be  formed  by  two  parabolas  as  shown  in  Fig.  57d. 

The  vertical  shear  modifies  in  practice  the  shape  of  these 
forms  near  their  ends.  For  instance,  a  cantilever  beam  loaded 
at  the  end  with  P  requires  a  section  area  at  the  end  equal  to 
P/S8  where  S8  is  the  allowable  shearing  unit-stress.  This  section 
area  should  continue  until  a  value  of  x  is  reached  where  the 
same  section  area  is  found  from  the  equation  of  the  form  of 
uniform  strength.  Exact  agreement  with  theoretic  conditions 
is  rarely  possible  on  account  of  the  expense  of  manufacture, 
and  in  fact  cast  iron  is  the  only  material  which  has  been  advan- 
tageously used  for  these  forms.  A  cantilever  of  structural  steel 
is  built  in  a  different  way  (Art.  58). 


FIG.  57 c.  FIG.  57d. 

The  deflection  of  a  cantilever  beam  of  uniform  strength  is 
evidently  greater  than  that  of  one  which  has  a  constant  cross- 
section  equal  to  the  greatest  cross-section  of  the  former,  since 
the  unit-stress  S  which  acts  only  at  the  wall  in  the  latter  case, 
acts  throughout  the  entire  length  in  the  former.  In  any  case 
it  may  be  determined  from  the  general  formula  El .  d2y/dx2  =  M 
by  substituting  for  M  and  /  their  values  in  terms  of  x,  integrating 
twice,  determining  the  constants,  and  then  making  x  equal  to  / 
for  the  maximum  value  of  y. 

For  a  cantilever  beam  loaded  at  the  end  and  of  constant 
breadth,  as  in  Fig.  57a,  this  formula  becomes, 


Integrating  twice  and  determining  the  constants,  as  in  Art.  54, 


146  CANTILEVER  AND  SIMPLE  BEAMS  CHAP.  VI 

the  equation  of  the  elastic  curve  is  found  to  be, 


In  this  let  x=l,  then  y  is  the  deflection  /  of  the  end,  and  /  = 
8>Pl3/Ebdi3,  which  is  double  the  deflection  of  a  cantilever  beam 
of  uniform  section  and  depth  d\. 

For  a  cantilever  beam  loaded  at  the  end  and   of   constant 
depth,  the  general  formula  becomes, 

I2PI 


_ 

8~*~~  Ebd?    Eb^ 

By  integrating  this  twice  and  determining  the  constants  as  before, 
the  equation  of  the  elastic  curve  is  found,  whence  the  deflection 
is  f=6Pl3/Ebid3  which  is  fifty  percent  greater  than  that  of  a 
cantilever  of  uniform  section  and  breadth  b\. 

Prob.  57.  A  cast-iron  cantilever  beam  of  uniform  strength  is  to  be 
4  feet  long,  3  inches  in  breadth,  and  to  carry  a  load  of  15  ooo  pounds 
at  the  end.  Find  the  proper  depths  for  every  foot  in  length,  using 
3  ooo  pounds  per  square  inch  for  the  horizontal  unit-stress,  and  4  ooo 
pounds  per  square  inch  for  the  shearing  unit-stress. 

ART.  58.     SIMPLE  BEAMS  OF  UNIFORM  STRENGTH 

In  the  same  manner  as  that  of  the  last  article  it  is  easy  to 
deduce  the  forms  of  uniform  strength  for  simple  beams  of  rectan- 
gular cross-section. 

For  a  load  at  the  middle  and  breadth  constant,  M  =  %Px, 
and  hence,  %Sbd2  =  \Px.  Accordingly  d2  =  (^P/Sb)xt  from  which 
values  of  d  may  be  found  for  assumed  values  of  x.  Here  the 
profile  of  the  beam  will  be  parabolic,  the  vertex  being  at  the 
support,  and  the  maximum  depth  under  the  load;  if  d\  is  the 
depth  at  the  middle,  the  equation  of  the  parabola  becomes 


For  a  load  at  the  middle  and  depth  constant,  M  =  \Px  as 
before.  Hence  b  =  (^P/Sd2)x)  and  the  plan  must  be  triangular 
or  lozenge-shaped,  the  width  uniformly  increasing  from  the 
support  to  the  load.  If  b\  is  the  breadth  at  the  middle,  the  equa- 
tion of  the  line  becomes  &=&i  (#/}/)» 


147 


ART.  58  SIMPLE    BEAMS    OF   UNIFORM    STRENGTH 


For  a  uniform  load  and  constant  breadth,  M  =  \wlx  — 
and  hence,  d2  =  (^w/Sb)(lx-x2),  and  the  profile  of  the  beam  must 
be  elliptical,  or  preferably  a  half-ellipse.  If  di  is  the  depth  at  the 
middle  the  equation  of  the  ellipse  becomes  d2  =  (^di2/l2)(lx  —  x2). 

For  a  uniform  load  and  constant  depth,  b=  (^w/Sd2)(lx  — x2), 
hence  the  plan  should  be  formed  of  two  parabolas  having  their 
vertices  at  the  middle  of  the  span.  If  bi  is  the  breadth  at  the 
middle  of  the  span,  this  equation  becomes  b  =  ($i/l2)(lx  —  x2). 

The  figures  for  these  four  cases  are  purposely  omitted,  in 
order  that  the  student  may  draw  them  for  himself;  if  any  dif- 
ficulty be  found  in  doing  this,  let  numerical  values  be  assigned 
to  the  constant  quantities  in  each  equation  and  the  variable 
breadth  or  depth  be  computed  for  different  values  of  x. 

In  the  same  manner  as  in  the  last  article,  it  can  be  shown 
that  the  deflection  of  a  simple  beam  of  uniform  strength  loaded 
at  the  middle  is  double  that  of  one  of  constant  cross-section  when 
the  breadth  is  constant,  and  is  one  and  one-half  times  as  much 
when  the  depth  is  constant. 


T== 

•fl 

J 

A                        B                                                      C                                             D" 
a                        b                                                       c                                            d 

Fig.  58 

Cast-iron  simple  beams  are  sometimes  made  approximately 
in  the  forms  required  by  the  above  equations,  care  being  taken 
to  provide  sufficient  sectional  area  near  the  supports  to  safely 
carry  the  vertical  shears ;  such  beams  are  mainly  used  in  machine- 
shops  on  planers  to  carry  the  cutting  tool.  Travelling  cranes 
used  in  shops  are  also  approximately  of  this  form,  but  these 
are  made  by  riveting  steel  plates  and  shapes  so  that  the  section 
areas  are  not  rectangular.  ' 

Plate  girders,  used  extensively  in  buildings  and  bridges,  are 
made  by  riveting  together  four  angles,  a  web  plate,  and  cover 
plates.  Fig.  58  shows  the  general  arrangement  without  the 


J.48  CANTILEVER  AND  SIMPLE  BEAMS  CHAP.  VI 

rivets.  The  section  at  A  is  made  ample  to  resist  the  shear  and 
that  at  B  to  resist  either  shear  or  moment;  there  being  no  cover 
plates  on  the  distance  AB.  Between  B  and  C  two  cover  plates 
are  used  to  provide  sufficient  section  for  the  moment  at  C,  while 
between  C  and  D  two  additional  cover  plates  are  added  so  that 
the  section  at  the  middle  D  will  be  sufficient  to  resist  the  maxi- 
mum moment  at  that  place.  A  plate  girder,  then,  is  approxi- 
mately a  beam  of  uniform  strength. 

Prob.  58a.  Draw  the  profile  for  a  cast-iron  simple  beam  of  uniform 
strength,  the  span  being  8  feet,  breadth  3  inches,  and  load  at  the  mid- 
dle 30  ooo  pounds;  using  the  same  working  unit-stresses  as  in  Prob.  57. 

Prob.  586.  Compute  the  deflection  of  a  steel  spring  of  constant 
depth  and  uniform  strength  which  is  6  inches  wide  at  the  middle,  52 
inches  long,  and  loaded  at  the  middle  with  600  pounds,  the  depths 
being  such  that  the  uniform  fiber  stress  is  20  ooo  pounds  per  square 
inch. 


ART.  59  BEAM  OVERHANGING  ONE  SUPPORT  149 

CHAPTER  VII 
OVERHANGING  AND  FIXED  BEAMS 

ART.  59.    BEAM  OVERHANGING  ONE  SUPPORT 

A  beam  is  said  to  be  'fixed  '  at  a  support  when  it  is  subject 
to  such  constraint  that  the  elastic  curve  is  there  horizontal.  The 
cantilevers  discussed  in  the  preceding  articles  have  been  fixed 
at  one  end  by  the  restraint  of  the  wall  and  the  maximum  moment 
has  been  found  to  occur  at  that  end.  Beams  are  sometimes 
fixed  at  one  end  and  supported  at  the  other  (Art.  60)  and  some- 
times fixed  at  both  ends  (Art.  62).  The  effect  of  this  restraint 
is  to  diminish  the  deflection,  and  hence  the  strength  and  stiff- 
ness are  usually  increased. 

Beams  overhanging  one  support,  as  in  the  following  figures, 
may  be  said  to  be  fixed  when  the  lengths  and  loads  have  such 
values  that  the  tangent  to  the  elastic  curve  at  that  support  is 
horizontal.  This  condition  is  rarely  fulfilled,  but  the  discussion 
of  overhanging  beams  is  very  useful  and  important.  A  canti- 
lever beam  has  its  upper  fibers  in  tension  and  the  lower  in  com- 
pression, while  a  simple  beam  has  its  upper  fibers  in  compression 
and  the  lower  in  tension.  Evidently  a  beam  overhanging  one 
support  has  its  overhanging  part  in  the  condition  of  a  canti- 
lever and  the  part  near  the  other  end  in  the  condition  of  a  simple 
beam.  Hence  there  must  be  a  point  where  the  curvature  changes 
from  positive  to  negative,  and  where  the  fiber  stresses  change 
from  tension  to  compression.  This  point  i  is  called  the  'Inflec- 
tion Point ' ;  it  is  the  point  where  the  bending  moment  is  zero, 
for  if  the  curvature  changes  from  positive  to  negative,  M  must 
do  likewise  (Art.  45) .  An  overhanging  beam  is  said  to  be  sub- 
ject to  a  constraint  at  the  support  beyond  which  the  beam  projects, 
or,  in  other  words,  there  is  a  stress  in  the  horizontal  fibers  over 
that  support. 

Since  the  beam  has  but  two  supports,  its  reactions  may  be 


150 


OVERHANGING  AND  FIXED  BEAMS 


CHAP.  VII 


found  by  using  the  principle  of  moments  as  in  Art.  36.  Thus, 
if  the  distance  between  the  supports  be  /,  the  length  of  the  over- 
hanging part  be  m,  and  the  uniform  load  per  linear  foot  be  w, 
the  two  reactions  for  Fig.  59a  are, 

RI  =  \wl — \wm  (m/l)  R2  =  \wl  +  wm + \wm  (m/l) 

From  these  the  vertical  shear  at  any  section  may  be  computed 
from  its  definition  in  Art.  37  and  the  bending  moment  from  its 
definition  in  Art.  38,  bearing  in  mind  that  for  a  section  beyond 
the  right  support  the  reaction  R2  must  be  considered  as  a  force 
acting  upward.  Thus,  for  any  section  distant  x  from  the  left 
support, 

When  x  is  less  than  /  iVhen  x  is  greater  than  / 

V=Ri-wx  V= 


i'he  curves  corresponding  to  these  equations  are  shown  on 
Fig.  590.  The  shear  curve  consists  of  two  straight  lines;  V  =  ft 
when  X  —  QJ  and  V  =  o  when  x  =  R\/w\  at  the  right  support 
V*=Ri  —  wl  from  the  first  equation;  V  =  Ri  +  R2—  wl  from  the 
second,  and  V  =  o  when  x  =  l+m.  The  moment  curve  consists 
of  two  parts  of  parabolas;  M  =  o  when  x  =  ot  and  M  is  a  maxi- 
.num  where  the  shear  passes  through  zero;  at  the  inflection 
point  M  =  o  and  x  =  2Ri/w;  also  M  has  its  maximum  negative 
value  at  the  right  support  where  the  shear  again  passes  through 
zero,  and  M  =  o  when  x=l+m.  The  diagrams  show  clearly 
the  distribution  of  shears  and  moments  throughout  the  beam. 

*^-O_4_-^-J 


Fig.  590  Fig.  59& 

In  any  particular  case  it  is  best  to  work  out  the  numerical 
values  without  using  the  above  algebraic  expressions.  For 
example,  if  /  =  2O  feet,  m  =  io  feet,  and  w  =  ^o  pounds  per  linear 


ART  59  BEAM    OVERHANGING    ONE   SUPPORT  151 

foot,  the  reactions  are  ^1=300  and  ^2  =  900  pounds.  Then  the 
point  of  zero  shear  or  maximum  moment  is  at  #  =  7.5  feet,  the 
inflection  point  at  #=15  feet,  the  maximum  shears  are  +300, 
—  500,  and  +400  pounds,  and  the  maximum  bending  moments 
are  +1125  and  --  2000  pound-feet.  Here  the  negative  bend- 
ing moment  at  the  right  support  is  numerically  greater  than 
the  maximum  positive  moment.  The  relative  values  of  the 
two  maximum  moments  depend  on  the  ratio  of  m  to  /;  if  w  =  o, 
there  is  no  overhanging  part  and  the  beam  is  a  simple  one;  if 
w  =  J/,  the  case  is  that  just  discussed;  if  m=-l,  the  reaction  RI 
is  zero,  and  each  part  is  a  cantilever  beam. 

After  having  thus  found  the  maximum  values  of  V  and  M 
the  beam  may  be  investigated  by  the  application  of  the  shear 
and  flexure  formulas  of  Art.  41  in  the  same  manner  as  a  canti- 
lever or  simple  beam.  By  the  use  of  formula  (45)  the  equation 
of  the  elastic  curve  between  the  two  supports  may  be  deduced 
by  two  integrations  and  the  proper  determination  of  the  con- 
stants, and  it  is, 

2$EIy = 4^1  (x3  -  I2x)  -w(x*-  I3x) 

From  this  the  maximum  deflection  for  any  particular  case  may 
be  determined  by  obtaining  the  derivative  of  y  with  respect  to  x\ 
equating  it  to  zero,  solving  for  x,  and  then  finding  the  corre- 
sponding value  of  y. 

If  concentrated  loads  be  placed  at  given  positions  on  the 
beam  the  reactions  are  found  by  the  principle  of  moments,  and 
then  the  entire  investigation  can  be  made  by  the  methods  above 
described.  Fig.  596  shows  the  shear  and  moment  diagram  for 
two  loads  and  here,  as  always,  the  maximum  moments  occur 
at  the  sections  where  the  shears  change  sign. 

Prob.  59a.  Three  men  carry  a  stick  of  timber,  one  taking  hold  at 
one  end  and  the  other  two  at  a  common  point.  Where  should  this 
point  be  so  that  each  may  bear  one-third  the  weight  ?  Draw  the  shear 
and  moment  diagrams. 

Prob.  59&.  A  beam  20  feet  long  has  one  support  at  the  right  end 
and  one  support  at  5  feet  from  the  left  end.  At  the  left  end  is  a  load 
of  1 80  pounds,  and  at  6  feet  from  the  right  end  is  a  load  of  125  pounds, 


152  OVERHANGING  AND  FIXED  BEAMS  CHAP,  vn- 

Find  the  reactions,  the  inflection  point,  and  draw  the  shear  and  moment 
diagrams. 

ART.  60.    BEAM  FIXED  AT  ONE  END 

A  beam  with  one  overhanging  end,  as  in  Fig.  59a,  has  the 
span  /  in  the  condition  of  a  beam  fixed  at  the  right  end  and  sup- 
ported at  the  other,  when  the  length  m  is  such  that  the  tangent 
to  the  elastic  curve  is  horizontal  over  the  right  support.  Fig.  600 
shows  the  practical  arrangement  of  such  a  beam,  the  right  end 
being  held  horizontal  by  the  restraint  of  the  wall.  The  usual 
arrangement  where  the  left  support  is  on  the  same  level  as  the 
lower  side  of  the  beam  at  the  wall  will  now  be  discussed,  the 
section  area  of  the  beam  being  constant  through  its  length. 

i 


Fig.  60a  Fig.  606 

Case  I.  Uniform  Load.  —  Let  R  be  the  reaction  of  the  left 
end  in  Fig.  60a,  and  x  the  distance  from  that  end  to  any  section. 
In  the  general  equation  of  the  elastic  curve  El  .  d2y/dx2  =  M, 
the  value  of  M  is  Rx-^wx2.  Integrating  once,  the  constant  is 
determined  from  the  condition  that  dy/dx  =  o  when  x  =  l.  Inte- 
grating again  the  constant  is  found  from  the  fact  that  y  =  o  when 
x  =  o;  then, 


Here  also  y=o  when  x=l,  and  therefore  R  =  %wl;    hence  the 
left  reaction  is  three-fourths  of  that  for  a  simple  beam. 

The  moment  at  any  point  now  is  M  =  %wlx  —  %wx2,  and  by 
placing  this  equal  to  zero,  it  is  seen  that  the  point  of  inflection 
is  at  #  =  |/.  The  maximum  positive  moment  occurs  when 
dM/dx=o  or  when  x  =  %l,  and  its  value  is  +ijb-w/2.  The  maxi- 
mum negative  moment  occurs  when  x  =  l  and  its  value  is  -\wl2- 


ART.  GO  BEAM  FIXED  AT  ONE  END  153 

The  distribution   of  shears   and  moments  is  as  shown  in  the 
diagrams. 

The  point  of  maximum  deflection  is  found  from  the  above 
equation  of  the  elastic  curve;  placing  the  derivative  equal  to 
zero  there  results  8#3  —  qlx2  +  13  =  o,  one  root  of  which  is 
x  =  0.42157,  while  the  others  are  inapplicable  to  this  problem. 
Hence  f=o.oo$4.wl4/EI  is  the  value  of  the  maximum  deflection. 

Case  II.  Load  P  at  Middle.  —  Here  it  is  necessary  to  con- 
sider that  there  are  two  elastic  curves  having  a  common  ordinate 
and  a  common  tangent  under  the  load,  since  the  expression, 
for  the  moment  are  different  on  opposite  sides  of  the  load.  Thus 
taking  the  origin  as  usual  at  the  supported  end, 

On  the  left  of  the  load, 

(a)       EI2  =  Rx  (b)       E/- 


(c) 
On  the  right  of  the  load  the  similar  equations  are, 


(b)'       E-= 
ooc 

(c)'  EIy=  %Rx*  -  %Px*+±Plx2+  C2x+  C4 

To  determine  the  constants  consider  in  (c)  that  y=o  when  x  =  o 
and   hence    that    C3=o.     In    (by    the    tangent    dy/dx  =  o   when 
x  =  l  and  hence  C2=—%Rl2.     Since  the  curves  have  a  common 
tangent  under  the  load,  (b)  =  (by  for  x  =  \l,  and  thus  the  value 
of  Ci  is  found.     Since  the  curves  have  a  common  ordinate  under 
the  load,  (c)  =  (c}'  when  x  =  \l,  and  thus  C4  is  found.     Then, 
(c)         24EIy=  4Rx*+  ^Pl2x-  1  2Rl2x 
(cy       4&EIy=  SRx*  -  SP*3+  1  2Plx2  -  2^Rl2x+  PI3 
are  the  equations  of  the   two   elastic  curves.     Making  x  =  l  in 
(c)f  the  value  of  y  is  zero,  and  then  the  left  reaction  is  R=^P. 

The  moment  on  the  left  of  the  load  is  now  M  =-?$Px,  and 
that  on  the  right  M=-\\Px  +  %Pl.  The  maximum  posi- 
tive moment  obtains  at  the  load  and  its  value  is  ^Pl.  The 


154  OVERHANGING  AND  FIXED  BEAMS  CHAP.  VII 

maximum  negative  moment  occurs  at  the  wall,  and  its  value 
is  -faPl.  The  inflection  point  is  at  x  =  fTl.  The  deflection 
under  the  load  is  readily  found  from  (c)  by  making  x  =  J/.  The 
maximum  deflection  occurs  at  a  less  value  of  x,  which  may  be 
found  by  equating  the  first  derivative  to  zero.  Fig.  606  shows  the 
distribution  of  shears  and  moments. 

Case  III.     Load  P  at  any  Point.  —  The  distance  of  the  load 
from  the  left  support  being  id  the  following  results  may  be  deduced 
by  a  method  exactly  similar  to  that  of  the  last  case. 
Reaction  at  supported  end       =  \P(?  —  3*:+  /c3) 
Reaction  at  fixed  end  =%P($K—  /c3) 

Maximum  positive  moment      =  \PlK.(z  —  3*+  K3) 
Maximum  negative  moment    =  JP/(/c—  K3) 

The  absolute  maximum  deflection  for  this  case  occurs  under  the 
load  when  x  =  0.4141,  and  its  value  will  be  found  to  be  given  by 


Prob.  60a.  Draw  the  shear  and  moment  diagrams  for  a  span  of  12 
feet,  due  to  a  load  P  at  10  feet  from  the  left  end. 

Prob.  60fr.  Find  the  position  of  load  P  which  gives  the  maximum 
positive  moment.  Find  also  the  position  which  gives  the  maximum 
negative  moment.  Compute  these  maximum  moments  and  compare 
them  with  those  due  to  a  load  at  the  middle. 

ART.  61.     BEAMS  OVERHANGING  BOTH  SUPPORTS 

When  a  beam  overhangs  both  supports,  the  moments  for 
sections  beyond  the  supports  are  negative,  and  in  general  between 
the  supports  there  will  be  two  inflection  points.  If  the  over- 
hanging lengths  are  equal,  the  reactions  will  be  equal  under 
uniform  load,  each  being  one-half  the  total  load.  In  any  case, 
whatever  be  the  kind  of  loading,  the  reactions  may  be  found  by 
the  principle  of  moments  (Art.  36),  and  then  the  vertical  shears 
and  bending  moments  may  be  deduced  for  all  sections,  after 
which  the  shear  and  flexure  formulas  (Art.  41)  can  be  used  for 
any  special  problem. 

Under  a  uniformly  distributed  load,  each  overhanging  end 
being  of  length  m,  and  the  middle  span  being  /,  each  reaction  is 


ART.  61 


BEAMS  OVERHANGING  BOTH  SUPPORTS 


155 


,  the  maximum  shears  at  the  supports  are  wm  and  % 
the  maximum  moment  at  the  middle  is  +  w(J/2-  Jw2),  the  maxi- 
mum moment  at  each  support  is  —  Jww2,  and  the  inflection 
points  are  distant  J(/2—  4W2)*  from  the  middle  of  the  beam. 
Fig.  61  a  shows  the  distribution  of  moments  for  this  case.  When 
m=o,  the  beam  is  a  simple  one ;  when  /  =  o,  it  consists  of  two 
cantilever  beams.  When  m  is  equal  to  or  greater  than  J/,  there 
are  no  positive  moments  in  the  middle  span. 


Fig.  61a  Fig.  616 

When  concentrated  loads  are  on  the  beam,  as  in  Fig.  616, 
the  reactions  are  readily  found  by  the  method  of  Art.  36,  the 
shears  and  moments  computed  for  several  sections  by  the  defini- 
tions of  Arts.  37  and  38,  and  the  shear  and  moment  diagrams 
may  then  be  drawn.  The  maximum  negative  moments  occur  at 
the  supports  and  the  maximum  positive  moment  under  one 
of  the  concentrated  loads.  The  final  maximum  shears  and 
moments  due  to  both  uniform  and  concentrated  loads  are  to  be 
obtained  by  combining  the  values  found  for  these  loads.  When 
the  concentrated  loads  are  light,  it  often  happens  that  the  final 
maximum  positive  moment  will  be  between  two  loads. 

Prob.  610.  For  Fig.  61a  find  the  ratio  of  /  to  m  in  order  that  the 
maximum  positive  moment  may  numerically  equal  the  maximum  nega- 
tive moment. 

Prob.  6lb.  A  beam  30  feet  long  has  one  support  at  5  feet  from  the 
left  end,  and  the  other  support  at  10  feet  from  the  right  end.  At  each 
end  there  is  a  load  of  156  pounds  and  half-way  between  the  supports 
there  is  a  load  of  344  pounds.  Construct  the  shear  and  moment 
diagrams. 

Prob.  61c.  For  Fig.  Qla  find  the  ratio  of  /  to  m  in  order  that 
there  may  be  no  positive  moment. 


156  OVERHANGING  AND  FIXED  BEAMS          CHAP.  Vll 


ART.  62.    BEAMS  FIXED  AT  BOTH  ENDS 

If,  in  Fig.  61  a,  the  distances  m  be  such  that  the  elastic  curve 
over  the  supports  is  horizontal,  the  central  span  /  is  said  to  be 
a  beam  fixed  at  both  ends.  The  length  m  which  will  cause  the 
beam  to  be  horizontal  at  the  support  can  be  determined  by  the 
help  of  the  elastic  curve.  For  uniform  load,  the  bending  moment 
at  any  section  in  the  span  /  distant  x  from  the  left  support  is, 

M=  (wm-\-  \wT)x 
which  may  be  written  in  the  simpler  form, 


where   MI   represents   the   unknown   bending   moment    — 

at  the  left  support.     The  distance  m  can  hence  be  found  when 

MI  has  been  determined. 

Again,  for  a  single  load  P  at  the  middle  of  /  in  Fig.  6  la,  the 
elastic  curve  can  be  regarded  as  kept  horizontal  at  the  left  sup- 
port by  a  load  Q  at  the  end  of  the  distance  m.  Then  the  bend- 
ing moment  at  any  section  distant  x  from  the  left  support,  and 
between  that  support  and  the  middle,  is, 

M=(Q+$P)x-Q(m+x)  or  M=Ml  +  %Px 

in  which  MI  denotes  the  unknown  moment  —  Qm  at  the  left 
support.  The  problem  of  finding  the  bending  moment  at  any 
section  hence  reduces  to  that  of  determining  MI  the  moment 
at  the  left  support, 

Case  I.  Uniform  Load.  —  For  this  case  the  differential 
equation  of  the  elastic  curve  becomes, 


Integrating  twice,  making  dy/dx  =  o  when  x  =  o  and  also  when 
x=l,  there  is  found  M\=  ~fawl?,axi.d  the  equation  of  the  elastic 

curve  is 

24EIy=  w(  -I2x2  +  2lx*  -x4) 

from  which  the  maximum  deflection  is  found  to  be  f=*-g%Twl4/EI. 
The  inflection  points  arejocated  by  placing  M  equal  to  zero, 
which  gives  x  =  ty(i±$\/3).  The  maximum  positive  moment 
is  at  the  middle  and  its  value  is  ^ivl2',  accordingly  the  horizontal 


ART.  62 


BEAMS  FIXED  AT  BOTH  ENDS 


157 


stress  upon  the  fibers  at  the  middle  of  the  beam  is  one-half 
that  at  the  ends.  The  vertical  shear  at  the  left  end  is  %wl,  at 
the  middle  o,  and  at  the  right  end  —  \wl.  Fig.  62a  shows  the 
shear  and  moment  diagrams. 

Case  II.  Load  P  at  Middle. — Here  the  differential  equation 
of  the  elastic  curve  is  El .  $2y/dx2=Mi  +  %Px  and  in  a  manner 
similar  to  that  of  the  last  case  it  is  easy  to  find  that  the  maxi- 
mum negative  moments  are  fP/,  that  the  maximum  positive 
moment  is  |P/,  that  the  inflection  points  are  half-way  between 
the  supports  and  the  load,  and  that  the  maximum  deflection  is 


Fig.  620 


Fig.  626 


Case  III.  Load  P  at  any  Point.— Let  the  load  P  be  at  the 
distance  id  from  the  left  end,  K  being  any  number  less  than  unity. 
Let  M\  and  R\  denote  the  unknown  bending  moment  and  reac- 
tion at  that  end.  Then  for  any  section  on  the  left  of  the  load 
M  =  M \-\-R\x,  and  for  any  section  on  the  right  of  the  load  M  = 
Mi  +  RiX  —  P(x  —  id).  By  inserting  these  in  the  differential 
equation  (45),  integrating  each  twice  and  establishing  sufficient 
conditions  to  determine  the  unknown  Ml  and  RI  and  also  the 
constants  of  integration,  the  following  results  may  be  deduced: 

Reaction  at  left  end     =  P(i  -3*2+  2/c3) 
Reaction  at  right  end  =  P/c2(3  — 2/c) 
Moment  at  left  end      =  —Plic(i  ~2K+  K2) 
Moment  at  right  end  =  —PlK2(i  -  K) 
Moment  under  load    =  +  P//c2(2  —4*+  2K2) 

When  K  =  %,  the  load  is  at  the  middle  and  these  results  reduce  to 
the  values  found  in  Case  II.  The  maximum  deflection  occurs 
under  the  load  when  it  is  at  the  middle. 


158  OVERHANGING  AND  FIXED  BEAMS          CHAP,  vii 

Prob.  62#.  From  the  above  results  for  Case  III  deduce  the  positions 
of  the  two  inflection  points.  Find  the  inflection  points  when  two 
equal  loads  are  at  \l  and  }/  from  the  left  end. 

Prob.  G2b.  What  medium-steel  I  beam  is  required  for  a  span  of 
24  feet  to  support  a  uniform  load  of  25  ooo  pounds  with  a  unit-stress 
S  of  1 6  ooo  pounds  per  square  inch,  the  ends  being  merely  supported? 
What  one  is  needed  when  the  ends  are  fixed  ? 


ART.  63.     COMPARISON  OF  BEAMS 

As  the  maximum  moments  for  fixed  beams  are  generally 
less  than  for  simple  beams,  their  strength  is  relatively  greater. 
This  was  to  be  expected,  since  the  constraint  lessens  the  deflec- 
tion which  would  otherwise  occur.  Under  a  uniform  load  W, 
the  maximum  moment  for  a  simple  beam  is  \Wl,  that  for  a  beam 
fixed  at  one  end  is  %Wl,  and  that  for  a  beam  fixed  at  both  ends 
is  TsWl',  the  fixing  of  one  end  does  not  increase  the  strength, 
but  the  fixing  of  both  ends  increases  it  fifty  percent. 

For  a  single  load  P  the  maximum  moment  for  a  simple  beam 
is  \Pl  and  that  for  a  beam  fixed  at  both  ends  is  \Pl\  hence  th' 
strength  of  the  latter  beam  under  a  concentrated  load  is  double 
that  of  the  former.  For  the  beam  fixed  at  one  end  and  sup- 
ported at  the  other,  it  may  be  shown  that  the  maximum  posi- 
tive moment  due  to  a  load  P  is  o.i74P/  and  that  the  maximum 
negative  moment  is  0.192 P/,  the  latter  occurring  when  the  load 
is  at  a  distance  o.tf'jl  from  the  supported  end;  hence  the  strength 
of  this  beam  is  intermediate  between  that  of  the  simple  beam 
and  that  of  the  beam  with  both  ends  fixed. 

With  respect  to  stiffness,  the  advantage  is  always  on  the  side 
of  beams  with  fixed  ends.  Under  uniform  load  the  deflection 
of  the  simple  beam  is  jfaWP/EI,  while  for  a  beam  fixed  at 
both  ends  it  is  one-fifth  of  this  amount;  hence  the  beam  fixed  at 
both  ends  is  five  times  as  stiff  as  the  simple  beam.  Under  a  single 
load  a  similar  comparison  of  the  deflections  shows  that  the  beam 
fixed  at  both  ends  is  four  times  as  stiff  as  the  simple  beam.  Foi 
the  beam  with  one  end  fixed,  the  degree  of  stiffness  is  interme- 
diate between  those  for  the  other  cases.  The  advantage  of 


ART.  64  SUPPORTS    ON    DIFFERENT   LEVELS  159 

fixing  the  ends  is  hence  much  greater  with  respect  to  stiffness 
than  with  respect  to  strength. 

Table  12,  at  the  end  of  this  volume,  recapitulates  the  above 
results,  and  also  those  deduced  in  Art.  56  for  cantilever  and 
simple  beams.  In  all  cases  W  represents  the  load  whether  con- 
centrated or  uniformly  distributed.  The  results  given  for  beams 
with  horizontal  restraint,  it  may  be  observed,  have  been  deduced 
from  the  equation  of  the  elastic  curve,  which  is  only  valid  when 
the  elastic  limit  of  the  material  is  not  exceeded  by  the  unit-stress  S 
at  the  dangerous  section.  When  beams  with  one  or  both  ends 
fixed  are  loaded  so  that  the  elastic  limit  is  exceeded,  the  above 
results  deduced  for  reactions,  moments,  and  deflections  are  not 
applicable  except  as  approximations. 

Let  i/a  represent  the  numerical  coefficient  in  the  column 
of  maximum  moments  in  Table  12  and  i//?  the  numerical  coeffi- 
cient in  the  column  of  maximum  deflections.  Then,  as  in  Art. 
56,  the  relation  between  the  unit-stress  S  and  the  deflection  / 

is  g'ven  by, 

S/f=pEc/aP  or  aSP=pEcf 

provided  the  elastic  limit  of  the  material  is  not  exceeded. 

Prob.  63a.  For  a  uniformly  loaded  beam  with  equal  overhanging 
ends,  derive  a  formula  for  the  deflection  at  the  middle. 

Prob.  63ft.  Find  the  deflection  of  a  9-inch  steel  I  beam  of  6  feet 
span  and  fixed  ends  when  loaded  at  the  middle  so  that  the  tensile  and 
compressive  stresses  at  the  dangerous  section  are  16  800  pounds  per 
square  inch. 

ART.  64.     SUPPORTS  ON  DIFFERENT  LEVELS 

For  all  beams  thus  far  discussed  the  two  supports  of  the 
beam  have  been  taken  as  in  the  same  horizontal  plane,  this  being 
the  usual  case  in  practice.  A  depression  of  one  support  below 
the  level  of  the  other  may,  however,  occur,  and  its  influence 
will  now  be  considered,  this  depression  being  taken  as  very  small 
so  that  the  length  of  any  small  part  of  the  elastic  curve  does  not 
sensibly  differ  from  that  of  its  horizontal  projection  (Art.  45). 

For  the   simple   beam  in   Fig.  64a,  let  the  left   support   be 


160  OVERHANGING  AND  FIXED  BEAMS          CHAP.  VII 

depressed  the  distance  h  below  the  level  of  the  right  support 
and  let  the  horizontal  distance  between  the  supports  be  /;  then 
h/l  is  the  tangent  of  the  angle  of  inclination,  or  tan  6=h/l. 
Now,  for  a  load  P  at  the  horizontal  distance  d  from  the  left 
support,  the  left  vertical  reaction  is  found  by  taking  an  axis  of 
moments  at  the  right  support  to  be  P(i  —  K)  which  is  independent 
of  6  and  the  same  as  if  the  supports  were  on  the  same  level.  The 
shears  and  moments  throughout  the  beam  due  to  the  vertical 
forces  are  hence  unaltered  by  the  inclination.  In  computing, 
the  fiber  unit-stress  from  the  flexure  formula  S .  I/c  =  M,  the 
dimensions  used  in  finding  I/c  are  those  of  a  normal  section  of 
the  inclined  beam,  and  hence  the  unit-stress  S  is  that  on  the 
inclined  upper  and  lower  surfaces  of  the  beam.  A  simple  beam 
is  therefore  unaffected  by  a  slight  change  in  the  relative  levels 
of  its  supports,  unless  it  be  from  the  influence  of  the  horizontal 
forces  which  must  come  into  play  at  the  supports  to  prevent 
sliding. 


Fig.  64a  Fig.  646  Fig.  64c 

Let  Fig.  646  represent  a  beam  supported  at  the  left  end  and 
fixed  horizontally  at  the  right  end,  the  vertical  distance  of  the 
left  below  the  right  end  being  h.  Under  a  uniform  load  of  w 
per  linear  unit,  the  moment  at  a  section  distant  x  from  the  left 
end  is  M=R\x  —  %wx2,  where  RI  is  the  unknown  reaction  at 
the  left  support,  and  x  is  the  distance  of  any  section  from  that 
support.  Inserting  this  in  the  differential  equation  (45)  and 
integrating,  the  constant  of  integration  is  found  by  the  condition 
that  dy/dx=o  when  x=l.  Integrating  again,  the  constant  is 
found  by  the  condition  that  y  =  o  when  x=o,  and, 


is  the  equation  of  the  elastic  curve.     In  this  y  becomes  h  when 
x  becomes  /,  and  accordingly  the  left  reaction  is, 


This  shows  that  the  value  of   R   depends  upon  the  difference 
of  level  of  the  two  supports.     When  h  =  o,  the  case  is  the  same 


ART.  64        SUPPORTS  ON  DIFFERENT  LEVELS  161 

as  Case  I  in  Art.  60  and  R\='^wl\  when  h  is  positive,  or  the 
fixed  end  higher  than  the  supported  one,  RI  is  less  than  %wl-} 
when  h  is  negative,  or  the  fixed  end  lower  than  the  supported 
one,  ^i  is  greater  than  %wl.  The  formula  also  shows  that  the 
value  of  ^i  depends  upon  the  kind  of  material  of  the  beam, 
since  E  is  the  modulus  of  elasticity  of  the  material,  and  upon 
the  size  and  shape  of  its  cross-section,  since  these  are  included 
in  the  moment  of  inertia  /. 

Let  Fig.  64c  represent  a  beam  with  both  ends  fixed,  and  let 
h  be  the  vertical  height  of  the  right  above  the  left  end.  Under  a 
uniform  load  let  the  unknown  reaction  at  the  left  support  be  RI 
and  the  unknown  moment  be  MI.  Then  the  differential  equa- 
tion of  the  elastic  curve  for  an  origin  at  the  left  support  is,  as 
in  Art.  62, 


Integrating  this  twice,   and  introducing    sufficient  conditions  to 

determine  the  constants  and  the  values  of  RI  and  MI,  there  result, 

RI  =  fal-  (i  2EI/P)h  MI  =  -Jjwl2+  (6EI/l2)h 

and  accordingly  these  values  depend  not  only  upon  the  difference 
in  level  of  the  supports  but  also  upon  the  dimensions  and  kind 
of  material  of  the  beam.  When  h=o,  the  values  of  RI  and  MI 
are  the  same  as  those  deduced  in  Art.  62.  The  reaction  R2 
at  the  right  end  of  the  beam  is  wl—R]_  and  the  moment  there 
is  found  from  M2  =  Mi+Ril  —  ^wl2\  accordingly, 

R2=%wl+  (i2EI/l3)h  M2=  —Tl2wl2-(6EI/l2)h 

which  show  that  at  the  higher  end  of  the  beam  the  reaction  is 
always  positive  and  the  moment  always  negative,  while  at  the 
other  end  they  may  be  positive  or  negative  depending  on  the 
value  of  h. 

Not  only  the  reactions  and  moments  at  the  supports  but 
also  the  shears  and  moments  throughout  the  beam  undergo  change 
when  one  support  is  lowered  below  the  other.  To  ascertain 
the  magnitude  of  these  changes,  take  a  lo-inch  I  beam  of  struc- 
tural steel  which  weighs  .  25  pounds  per  linear  foot,  for  which 
£=30000000  pounds  per  square  inch  and  I  =122.1  inches4 


162  OVERHANGING  AND  FIXED  BEAMS          CHAP.  VII 

(Table  6).  Let  one  end  be  supported  and  the  other  fixed,  as 
in  Fig.  646  the  clear  span  be  15  feet  and  the  total  uniform  load 
W  be  1 6  400  pounds.  When  both  supports  are  on  the  same 
level,  the  left  reaction  is  %W=6  150  pounds,  and  the  maximum 
moment  is  4^=30750  pound-feet  =  369  ooo  pound-inches;  then 
the  flexure  formula  gives  5=369000/24.42=15  100  pounds  per 
square  inch,  which  is  a  satisfactory  unit-stress  for  steady  load. 
Now  let  the  left  support  be  lowered  1.2  inches  below  the  fixed 
end;  the  reaction  at  that  end  is  then  less  than  6  150  pounds  by 
the  amount  ^Elh/l3  =  3X30  ooo  000X122.1  Xi.2/i8o3=  2  260 
pounds,  so  that  it  is  ^1  =  6150-2260=3890  pounds.  Froni 
this  the  moment  at  the  fixed  end  is  found  to  be  M=RJ  —  ^Wl= 
775  800  pound-inches,  and  the  flexure  formula  then  gives  5^ 
31  800  pounds  per  square  inch,  which  is  too  high  a  value  for 
safety  as  it  is  but  little  less  than  the  elastic  limit  of  the  material. 

The  conclusion  of  this  investigation  is  that  beams  with  one 
or  both  ends  fixed  should  not  be  used  in  circumstances  where 
any  material  alteration  in  the  levels  of  the  supports  may  occur. 
The  above  formulas  show  that  when  the  left  support  in  Fig.  646 
is  lowered  the  distance  h=  W13/SEI  below  the  level  of  the  other, 
the  left  reaction  under  uniform  load  is  zero  and  the  beam  is  a 
cantilever  fixed  at  the  right  end.  The  same  condition  obtains 
for  the  beam  in  Fig.  64c  when  it  is  uniformly  loaded  and  the 
left  end  is  lowered  the  distance  Wfi/2^EI\  when  greater  depres- 
sions occur,  the  left  reaction  is  negative  and  the  beams  are  in 
the  condition  of  constrained  cantilevers  uniformly  loaded  and 
having  a  concentrated  load  at  the  free  end  (Art.  65).  It  should 
also  be  noted  that  the  moments  and  reactions  deduced  above 
are  only  valid  when  the  elastic  limit  of  the  materials  is  not 
exceeded. 

Prob.  640.  For  the  case  of  Fig.  64c,  deduce  expressions  for  the  dis- 
tance of  the  inflection  points  from  the  left  support. 

Prob.  646.  A  wooden  joist,  3X12X75  inches,  is  fixed  in  walls  at 
its  ends  and  carries  a  total  uniform  load  of  7  680  pounds.  Compute 
the  maximum  unit-stress  S  when  the  supports  are  on  the  same  level; 
also  when  one  support  is  lowered  one-quarter  of  an  inch  below  the 
other. 


ART.  65 


CANTILEVER   WITH    CONSTRAINT 


163 


ART.  65.     CANTILEVER  WITH  CONSTRAINT 

A  case  which  occurs  in  the  portal  of  a  bridge  is  shown  in 
Fig.  650,  where  AC  and  ac  represent  the  two  end  posts  connected 
by  bracing  which  transfers  the  wind  pressures  P  to  the  points 
B  and  b.  The  parts  BC  and  be  of  these  posts  are  cantilever 
beams  fixed  at  the  lower  ends  C  and  c,  while  they  are  also  kept 
fixed  at  B  and  b  by  the  restraint  of  the  portal  bracing.  By  turn- 
ing this  figure  clockwise  through  a  right  angle,  it  will  be  in  the 
position  of  the  end  view  of  a  bridge  portal. 


Fig.  65<z  Fig.  656 

To  investigate  a  cantilever  beam  in  which  the  free  end  is 
kept  horizontal  by  a  restraint  during  its  deflection,  let  Fig.  656 
be  considered.  Taking  the  origin  of  coordinates  at  the  free  end 
where  the  unknown  moment  is  MI,  the  differential  equation 
of  the  elastic  curve  is  El .  d2y/dx2-=M\  —  Px  for  any  section 
distant  x  from  that  end.  Integrating  this  it  becomes  El .  dy/dx  = 
MiX  —  ^Px2)  the  constant  being  zero  because  the  tangent  to  the 
elastic  curve  is  horizontal  at  the  left  end.  At  the  right  end,  for 
which  x=l,  the  tangent  is  also  horizontal,  and  hence  M \  =  +  \PL 
The  moment  at  the  right  end  now  is  M^=  +  ?Pl—Pl  =  —$Pl,  the 
inflection  point  is  at  #=J/,  since  M=%Pl—Px=o  locates  this 
point,  and  the  moment  diagram  can  then  be  drawn  as  shown. 

It  thus,  appears  that  the  maximum  moments  for  this  case  is 
only  one-half  as  large  as  the  common  cantilever  beam  loaded 
at  the  end.  The  deflection  of  the  end  may  be  found  by  inte- 
grating the  above  first  derivative,  and  it  is  f=Pl3/i2EI,  which  is 
only  one-fourth  of  that  of  the  common  cantilever  with  a  load 
at  the  end. 

Another  case  of  a  cantilever  with  restraint  at  the  free  end 
is  that  of  a  crank-pin  connected  by  two  webs  through  which 
the  torsion  of  a  shaft  is  transmitted.  A  figure  showing  such  a 


164  OVERHANGING  AND  FIXED  BEAMS          CHAP.  VII 

crank-pin  may  be  found  in  Art.  98,  and  it  is  there  seen  that  the 
bending  moments  due  to  the  transmitted  lateral  force  from  the 
web  are  exactly  the  same  as  those  deduced  above. 

A  cantilever  may  be  fixed  in  a  wall  at  any  angle  with  the 
horizontal.  In  that  event,  let  /  be  the  length  of  the  inclined 
beam,  and  let  the  axis  of  abscissas  be  taken  as  coinciding  with 
the  tangent  drawn  to  the  elastic  curve  at  the  wall;  then  if  the 
angle  of  inclination  be  small  the  formulas  deduced  in  Art.  54 
will  approximately  give  the  vertical  deflection  of  the  end  from 
that  tangent.  The  bending  moment  at  the  wall  is  found,  how- 
ever, by  using  the  horizontal  lever  arms  of  the  vertical  forces. 
When  the  angle  of  inclination  is  not  small,  the  cantilever  is  sub- 
ject to  combined  axial  stress  and  flexure  (Art.  101). 

Prob.  65a.  Deduce  the  moments  at  the  supports  for  Fig.  Q5b  with- 
out using  the  equation  of  the  elastic  curve. 

Prob.  656.  A  steel  crank-pin,  like  that  of  Fig.  98,  is  hollow,  18 
inches  in  outside  and  6  inches  in  inside  diameter  and  12  inches  in 
length  between  the  webs  into  which  it  is  fixed.  Compute  the  deflec- 
tion of  one  end  with  respect  to  the  ether  when  the  force  P  is  126  100 
pounds. 

ART.  66.     SPECIAL  DISCUSSIONS 

When  a  simple  beam  deflects  the  upper  side  is  shortened 
and  the  lower  side  elongated.  These  changes  in  length  are  due 
to  the  horizontal  compressive  and  tensile  stresses  acting  along 
those  sides,  and  the  amount  of  the  same  will  now  be  found. 
Let  S  be  the  unit-stress  acting  on  the  side  at  the  distance  c  from 
the  neutral  surface,  and  E  the  modulus  of  elasticity  of  the  mate- 
rial. Then,  from  Art.  9,  the  change  of  length  which  occurs  in 
the  distance  dx  is  (S/E)dx,  and  hence  the  total  change  of  length 
in  the  span  /  is  the  sum  of  all  these  elementary  values;  to  find 
this  sum  5  is  to  be  expressed  in  terms  of  x,  since  5  varies  through- 
out the  span.  From  the  flexure  formula  (41)  the  value  of  S 
is  Me/ 1,  and  for  a  simple  beam  under  the  uniform  load  w  per 
linear  unit  M  is  \wlx-  %wx2.  Accordingly  the  entire  change  of 
length  of  the  upper  or  lower  surface  of  the  beam  is, 


ART.  66  SPECIAL   DISCUSSIONS  165 

in  which  the  integration  is  to  be  made  between  the  limits  /  and  o. 
This  gives  e  =  wcl3/i2EI  which  is  the  amount  of  shortening  of 
the  upper  side  of  the  simple  beam  if  c  is  the  distance  of  that 
side  from  the  neutral  surface,  or  the  amount  of  elongation  of 
the  lower  side  of  the  beam  if  c  is  its  distance  from  the  neutral 
surface.  For  example,  let  the  beam  be  an  unloaded  steel  bar, 
2X2X72  inches,  then  the  formula  gives  e  =  o.ooo  0122  inches. 

Steel  bars  are  sometimes  used  in  measuring  the  base  lines 
of  geodetic  triangulations.  The  upper  diagram  of  Fig.  66#  shows 
such  a  bar  laid  upon  a  horizontal  plane  and  the  positions  of 
two  marks  upon  its  upper  surface  very  near  its  ends;  the  dis- 
tance between  these  marks  is  determined  with  precision  by  com- 
parisons with  a  standard.  When  the  bar  is  used  in  the  field, 
it  is  laid  upon  two  supports,  as  in  the  lower  diagram  of  Fig.  660, 
and  these  supports  should  be  so  placed  that  the  distance  between 
the  marks  remains  unaltered  by  the  deflection  of  the  beam  under 
its  own  weight.  In  this  beam  with  equal  overhanging  ends, 
the  upper  fibers  are  shortened  between  the  two  inflection  points 
and  elongated  elsewhere;  hence  it  is  possible  to  place  the  sup- 
ports at  such  a  distance  apart  that  the  amounts  of  shortening 
and  elongation  are  equal  and  then  the  distance  between  the 
two  marks  will  be  unaltered  by  the  bending  of  the  beam. 


««_ ,^4U          A?*" 


/  \ 

/  \ 


b— 


Fig.  66a  Fig.  666 

Let  L  be  the  length  of  the  beam,  which  is  practically  equal 
to  the  distance  between  the  two  marks,  m  the  length  of  each 
overhanging  end,  and  /  the  distance  between  the  supports;  thus 
2m  +  l=L,  and  if  the  ratio  of  /  to  m  is  found,  that  of  m  to  L  will 
be  known.  Let  x  be  any  distance  from  the  left  end,  then  the 
elongation  of  any  upper  fiber  on  the  overhanging  end  in  the 
distance  dx  is  (S/E)dx,  which  by  the  flexure  formula  becomes 
(Me /El) doc  or  (wc/2EI)x2dx.  Integrating  this  between  the 
limits  m  and  o,  there  results  e\  =  wcm^/6EI  for  the  elongation 


166  OVERHANGING  AND  FIXED  BEAMS  CHAP.  VII 

of  the  upper  fiber  of  the  overhanging  end.  Again,  the  moment 
M  at  a  section  in  the  central  span  distant  x  from  the  left  support 
is  RIX  —  \w(m  +  x)2,  in  which  RI  is  the  reaction  wm  +  ^wl.  It 
hence  follows  that, 

e2=%(wc/EI)f(lx-x2-m2)dx 

is  the  shortening  of  the  upper  fiber  in  the  central  span;  when 
this  is  integrated  between  the  limits  /  and  o  it  gives  62  = 
(wc/i2EI)(l3  —  6m2l).  Now  the  condition  that  there  shall  be 
no  change  of  length  in  the  upper  fiber  is  e2  —  261  =  0,  which 
leads  to  the  cubic  equation  I3  —  6m2l— 4W3  =  o.  This  equation 
gives  //w=2.732  and  the  two  other  roots  are  negative.  Hence 
m  =  0.2 1 13!,  and  /  =  0.5  7  74!,,  so  that,  if  the  length  of  the  bar 
\s  one  meter,  the  distances  /  and  m  should  be  577.4  and 
211.3  millimeters. 

Beams  of  triangular  section  are  rarely  or  never  used  in  prac- 
tice, since,  if  the  vertex  be  upward  the  load  cannot  safely  rest 
on  the  sharp  edge,  and  if  the  vertex  be  downward  it  cannot  safely 
rest  on  the  supports.  If  b  is  the  base  and  d  the  altitude,  as  in 
Fig.  666,  the  section  factor  I/c  is,  from  Arts.  42  and  43,  found 
to  be  -fabd2  and  hence,  from  Art.  56,  the  triangular  beam  is  one- 
quarter  as  strong  as  a  rectangular  one  having  the  same  breadth 
and  depth.  If  the  triangle  is  cut  off  at  a  depth  y  below  the  vertex, 
thus  forming  a  trapezoidal  section  with  the  bases  b  and  by/d 
and  the  height  d—y,  the  section  factor  with  respect  to  the  new 
neutral  axis  will  be  greater  than  -fabd2  when  the  value  of  y  is 
less  than  a  certain  limit.  Hence  the  strength  of  a  triangular 
section  may  be  increased  by  cutting  off  the  vertex.  This  case 
is  one  of  mere  theoretical  interest  and  will  not  be  developed 
here,  but  it  is  shown  on  page  181  of  Wood's  Resistance  of 
Materials  (New  York,  1882)  that  the  maximum  strength  of  the 
trapezoid  thus  formed  is  9  percent  greater  than  the  strength  of 
the  triangular  section  and  that  this  occurs  when  the  distance  y 
is  13  percent  of  the  altitude  d. 

The  following  are  a  few  interesting  problems  regarding  beams 
which  involve  ideas  that  have  not  received  detailed  discussion 
in  the  preceding  pages. 


ART.  66  SPECIAL   DISCUSSIONS  167 

Prob.  66<z.  Find  the  thickness  of  a  white-pine  plank  of  8  feet  span 
so  that  it  shall  not  bend  more  than  one-fifth  of  an  inch  under  a  head 
of  water  of  10  feet. 

Prob.  666.  Prove  that  the  greatest  possible  length  of  a  simple  beam 
oi  breadth  b,  depth  d,  for  an  assigned  flexural  unit-stress  S  is  (^Sd/^v)*, 
where  v  is  the  weight  of  a  cubic  unit  of  the  material. 

Prob.  66c.  A  simple  wooden  beam,  one  inch  square  and  15  inches 
long,  is  uniformly  loaded  with  100  pounds.  Find  the  angle  of  incli- 
nation of  the  elastic  curve  at  the  supports. 

Prob.  QQd.  A  simple  beam  of  structural  steel,  one  inch  in  diameter, 
is  of  such  a  length  that  the  flexural  unit-stress  at  the  middle,  due  to  its 
own  weight,  is  equal  to  the  elastic  limit.  Compute  the  inclination  of 
the  elastic  curve  at  the  supports. 

Prob.  66e.  A  rolled  steel  beam,  5^  meters  long  and  30  centimeters 
deep  (Table  13),  is  fixed  at  its  ends  and  carries  a  uniform  load  of  2  600 
kilograms.  Compute  the  greatest  horizontal  unit-stress  5  in  kilo- 
grams per  square  centimeter,  and  ascertain  the  factor  of  safety  of  the 
beam. 


168  CONTINUOUS  BEAMS  CHAP.  VIH 


CHAPTER  VIII 
CONTINUOUS  BEAMS 

ART.  67.     GENERAL  PRINCIPLES 

A  continuous  beam  is  one  resting  upon  several  supports,, 
these  being  usually  in  the  same  horizontal  plane.  A  simple 
beam  may  be  regarded  as  a  particular  case  of  a  continuous  beam 
where  the  number  of  supports  is  two.  The  ends  of  a  continuous 
beam  are  said  to  be  free  when  they  overhang,  supported  when 
they  merely  rest  on  abutments,  and  fixed  when  they  are  kept 
horizontal  by  the  restraint  of  walls;  the  most  common  cases 
are  those  of  supported  ends  and  this  chapter  will  be  mainly 
devoted  to  their  discussion. 

The  general  principles  of  the  preceding  chapters  hold  good 
for  all  kinds  of  beams.  If  a  vertical  plane  is  imagined  to  cut  any 
beam  at  any  point,  the  laws  of  Arts.  39  and  40  apply  to  the  stresses 
in  that  section.  The  resisting  shear  and  the  resisting  moment 
for  that  section  have  the  values  deduced  in  Art.  41  and  the  shear 
and  flexure  formulas  are, 

Ssa=V  S.I/c=M 

Here  S8  is  the  vertical  shearing  unit-stress  in  the  section,  and 
S  is  the  horizontal  tensile  or  compressive  unit-stress  on  the  fiber 
most  remote  from  the  neutral  surface,  c  is  the  shortest  distance 
between  that  fiber  and  neutral  surface  (Art.  42) ;  a  is  the  area 
of  the  cross-section  and  I  its  moment  of  inertia  with  repect  to 
the  neutral  axis.  V  is  the  vertical  shear  of  the  external  forces 
on  the  left  of  the  section,  and  M  is  the  bending  moment  of  those 
forces  with  reference  to  a  point  in  the  section.  For  any  given 
beam,  evidently  Ss  and  S  may  be  found  for  any  section  as  soon 
as  V  and  M  are  known,  and  these  are  determined  for  any  given 
loads  from  the  definitions  in  Arts.  37  and  38.  For  brevity  V 
and  M  will  hereafter  be  called  shear  and  moment. 

The  general  equation  of  the  elastic  line,  deduced  in  Art.  45, 


ART.  67  GENERAL   PRINCIPLES  169 

is  also  valid  for  all  kinds  of  beams.  It  is  EId2y/dx2  =  M,  where 
x  is  the  abscissa  and  y  the  ordinate  of  any  point  of  the  elastic 
curve,  and  E  the  coefficient  of  elasticity  of  the  material. 

The  shear  V  is  the  algebraic  sum  of  the  external  forces  on 
the  left  of  the  section,  or,  as  in  Art.  37, 

V  =  Reactions  on  left  of  section  minus  loads  on  left  of  section. 

For  simple  beams  and  cantilevers  the  determination  of  V  for 
any  special  case  was  easy,  as  the  left  reaction  could  be  readily 
found  for  any  given  loads.  For  continuous  beams,  however, 
it  is  not,  in  general,  easy  to  find  the  reactions,  and  hence  a  different 
method  of  determining  V  is  usually  necessary.  Let  Fig.  67 
represent  one  span  of  a  continuous  beam.  Let  V  be  the  shear 
for  any  section  at  the  |«—  —  x—  — 

distance   x  from   the  left  k—  Ki—  -* 


support,  and  V  the  shear     I  -  L  _  \ 

Af   ,  1  TZS 

at  a  section  infinitely  near 

to  the  left  support.     Also 

let   IP\   denote  the  sum 

of  all  the  concentrated  loads  on  the  distance  x}  and  wx  the  uniform 

load.     Then  because  V  is  the    algebraic   sum  of  all  the  vertical 

forces  on  its  left,  the  definition  of  vertical  shear  gives, 

V=V'-wx-IPl  (67) 

Hence  the  shear  V  can  be  determined  for  any  section  in  the  span 
as  soon  as  V  is  known. 

The  moment  M  is  the  algebraic  sum  of  the  moments  of  the 
external  forces  on  the  left  of  the  section  with  reference  to  a  point 
in  that  section,  or,  as  in  Art.  38, 

M=  moments  of  reactions  minus  moments  of  loads 
For  the  reason  just  mentioned,  it  is  in  general  necessary  to  deter- 
mine M  for  continuous  beams  by  a  different  method.  Let  Mf 
denote  the  moment  at  the  left  support  of  any  span  as  in  Fig.  67, 
and  M"  that  at  the  right  support,  while  M  is  the  moment  for 
any  section  distant  x  from  the  left  support.  Let  PI  be  any  con- 
centrated load  upon  the  space  x  at  a  distance  d  from  the  left 
support,  K  being  a  fraction  less  than  unity,  and  let  w  be  the  uniform 
load  per  linear  unit.  Since  the  shear  V  in  Fig.  67  is  equal  to 


170  CONTINUOUS  BEAMS  CHAP.  Vlli 

the  resultant  of  all  the  vertical  forces  on  the  left  of  a  section  just 
at  the  right  of  the  left  support,  let  m  be  the  distance  of  the  line 
of  action  of  that  resultant  to  the  left  of  that  support.  Then  the 
definition  gives,  for  the  moment  at  any  section, 

M=  V'(m+x)-wx  .  \x-I  .  Pi(x-id) 

But  the  quantity  V'm  is  equal  to  the  sum  of  the  moments  of 
all  the  forces  on  the  left  of  the  left  support  with  respect  to  that 
support  and  hence  it  is  the  moment  Mr  at  the  left  support  of 
the  span.  Hence, 

M=M'+  V'x-fyJoP-2  .  PI  (*-*/)  (67)' 

from  which  the  moment  M  may  be  found  for  any  section  in  the 
span  as  soon  as  M'  and  V  are  known. 

The  shear  V  at  the  support  of  the  span  may  be  easily  found 
:f  the  moments  Mr  and  M"  be  known.  Thus  in  equation  (67)' 
make  x=l,  then  M  becomes  M" ,  and  hence, 

V'l=M"-M'+  lwP  +  IPi(l—id)  (67)" 

and  hence  the  problem  of  the  discussion  of  continuous  beams 
consists  in  the  determination  of  the  moments  at  the  supports. 
When  these  are  known,  the  values  of  M  and  V  may  be  deter- 
mined for  every  section  in  any  span,  and  the  investigation  of 
questions  of  strength  and  deflection  be  made  from  the  formulas 
(41)  and  (45).  The  above  formulas  apply  to  cantilever  and 
simple  beams  also.  For  a  simple  beam,  M'  =  M"  =  o,  and 
Vf  =  R.  For  a  cantilever  beam,  Mf  =  o  for  the  free  end,  and 
M"  is  the  moment  at  the  wall. 

The  relation  between  the  moment  and  the  shear  at  any  sec- 
tion is  interesting  and  important.  At  a  section  distant  x  from 
any  support,  the  moment  is  M  and  the  shear  is  V.  At  the  sec- 
tion distant  x  +  dx  from  the  support,  the  moment  is  M  +  Vdx, 
which  may  also  be  expressed  as  M  +  8M.  Accordingly, 

Vdx=dM  or  dM/dx=V 

This  may  also  be  found  by  finding  the  derivative  of  M  with 

respect  to  x  from  (67)'  and  comparing  it  with  (67).     Therefore 

The  derivative  of  the  moment  equals  the  shear 


ART.  68  METHOD    OF    DISCUSSION  171 

and    from    this  it   is  seen   that  the    maximum  moments  occur 
at  the  sections  where  the  shear  passes  through  zero. 

Prob.  67.  A  bar  of  length  2/,  and  weighing  w  per  linear  unit,  is 
supported  at  the  middle.  From  (67)  and  (67)'  find  general  expressions 
for  the  shear  and  moment  at  any  section  on  the  left  of  the  support 
and  also  at  any  section  on  the  right  of  the  support.  Draw  the  shear 
and  moment  diagrams. 

ART.  68.     METHOD  OF  DISCUSSION 

The  theory  of  continuous  beam?  presented  in  the  following 
pages  includes  only  those  with  constant  cross-section  having 
the  supports  on  the  same  level,  since  only  such  are  used  in  engi- 
neering constructions.  Unless  otherwise  stated,  the  ends  will 
be  supposed  simply  to  rest  upon  their  supports,  so  that  there 
can  be  no  moments  at  those  points.  Then  die  end  spans  are 
somewhat  in  the  condition  of  a  simple  beam  with  one  overhang- 
ing end,  while  the  other  spans  are  somewhat  in  the  condition 
of  a  beam  with  two  overhanging  ends.  At  each  intermediate 
support  there  is  a  negative  moment,  and  the  distribution  of 
jhears  and  moments  due  to  uniform  load  is  that  shown  in  Fig.  68. 
When  an  end  span  is  short,  the  reaction  at  that  end  may  become 
zero  or  even  be  negative;  in  order  that  a  negative  reaction  may 
exist,  it  is  necessary  that  the  end  of  the  beam  be  anchored  or 
fastened  to  the  support. 


Fig.  68 

As  shown  in  Art.  67,  the  investigation  of  a  continuous  beam 
depends  upon  the  determination  of  the  moments  at  the  sup- 
ports. In  the  case  of  Fig.  68,  the  moments  at  the  supports  2, 
3,  and  4,  may  be  designated  M2,  MS,  and  M\.  Let  FI,  ¥2, 
FS,  and  F4  denote  the  shears  at  the  right  of  those  supports. 
The  first  step  is  to  find  the  moments  M2,  MS,  and  M±.  Then 
from  formula  (67)"  the  values  of  FI,  ¥2,  Vs,  and  V±  are  found, 


172  CONTINUOUS  BEAMS  CHAP  vill 

and  thus  by  formula  (67)'  an  expression  for  the  moment  in 
each  span  may  be  written,  from  which  the  maximum  positive 
moments  may  be  determined.  Lastly,  by  the  shear  and  flexure 
formulas  of  Art.  41  the  beam  may  be  investigated. 

For  example,  let  the  beam  in  Fig.  68  be  regarded  as  of  four 
equal  spans  and  uniformly  loaded  with  w  pounds  per  linear 
unit.  By  a  method  to  be  explained  in  the  following  articles  it 
may  be  shown  that  the  moments  at  the  supports  are, 

M2=  —fiwt2  M3=  -  ^wl2  M4=  -&wP 

From  formula  (67)"  the  shears  at  the  right  of  the  several  sup- 
ports are  found  to  have  the  values, 


And  from  (67)  those  on  the  left  of  the  supports  2,  3,  4,  5,  are 
found  to  be,  —  Jfw/,  —%$wl,  —%%wl,  —  JJw/.  From  formula 
(67)'  the  general  expressions  for  the  moments  now  are, 

for  first  span  ,  M  =  +  Jf  wlx  — 

for  second  span,  M=  — 

for  third  span,  M  =  — 

for  fourth  span,  M=—  ^  ivT2  +  $wlx  — 

From  each  of  these  equations  the  inflection  points  may  be  found 
by  putting  M  =  ot  and  the  section  of  maximum  positive  moment 
by  putting  dM/dx=o.  The  maximum  positive  moments  are 
found  to  have  the  following  values  : 


For  any  particular  case  the  beam  may  now  be  investigated  by 
the  use  of  the  shear  and  moment  formulas.  It  is  seen  that  the 
greatest  moment  is  that  at  support  2,  and  hence  this  need  only 
be  used  in  the  flexure  formulas. 

The  reactions  at  the  supports  are  readily  found  from  the 
values  of  the  adjacent  shears.  Thus,  for  the  above  case  RI=* 
Vi  =  fywl,  and  R2=  J|w/  +  Jfw/=ffw/.  But  perhaps  a  more 
satisfactory  method  will  be  to  find  them  directly  from  the  equation 
of  moments.  Thus  Ril—^wl2  =  M2,  whence  Ri  =  ^wl.  Again 
RiX2l+R2l  —  2wl2  =  M3,  whence  R2  =  %%wl.  From  the  symmetry 
of  the  spans  and  loads,  it  is  plain  that  R&  =  Ri  and  ^4  =  ^2- 


ART.  69  THEOREM  OF  THREE  MOMENTS  173 

The  equation  of  the  elastic  curve  in  any  span  is  found  by 
inserting  the  expression  for  M  in  El  .  §2y/dx2  =  M,  and  inte- 
grating twice.  In  general,  the  maximum  deflection  in  any  span 
will  be  found  intermediate  in  value  between  those  of  a  simple 
beam  and  one  fixed  at  its  ends. 

Prob.  68.  In  a  continuous  beam  of  three  equal  spans  the  negative 
bending  moments  at  the  supports  are  j$wl2.  Find  the  inflection  point, 
the  maximum  positive  moments,  and  the  reactions  of  the  supports. 

ART.  69.     THEOREM  OF  THREE  MOMENTS 

Let  the  figure  represent  any  two  adjacent  spans  of  a  continu- 
ous beam  having  the  lengths  /'  and  I"  and  the  uniform  loads 
itf  and  TV"  per  linear  unit.  Let  Mr  ,  M"  ,  and  Mfn  represent 
the  three  unknown  moments  at  the  supports.  Let  V  and  V" 
be  the  vertical  shears  at  the  right  of  the  first  and  second  sup- 
ports. Then,  for  any  section  distant  x  from  the  left  support 
in  the  first  span,  the  moment  is,  M  =  M'  +  V'x  —  \wx2.  Let  this 
be  inserted  in  the  general  formula  of  the  elastic  curve.  Integrating 
twice  and  determining  the  constants  by  the  conditions  that 
y  =  o  when  x=o  and  also  when  x=l'  the  tangent  dy/dx  of  the 
angle  which  the  tangent  to  the  elastic  curve  at  any  section  in 
the  first  span  makes  with  the  horizontal  is  found  to  be  given  by, 

24EI(dy/dx)  =  1  2M'(2X-V)  +  4F'(3*2  -I'2)  -  w'(4*3  -/'3) 

Similarly  if  the  origin  is  taken  at  the  next  support,  the  tangent 
of  inclination  at  any  point  in  the  second  span  is, 


The  two  curves  must  have  a  common  tangent  at  the  support 
where  they  meet,  in  order  that  the  beam  may  be  continuous 
Hence  make  x  =  V  in  the  first  equation  and  x=o  in  the  second, 
and  equate  the  results,  giving 

i2M'l'+8V'l'2-3w'l'3=-i2M'fl"-4V"l"2+w"l"3 
Now  let  V  and  V"  be  expressed  by  means  of  (67)"  in  terms  of 
M',  Mn  ',  and  Mm;  then  this  equation  reduces  to 

M'l'+2M"(l'+l")  +  M'"l"=  -\w'V*-\w"l"*  (69) 

which  is  called  the  theorem  of  three  moments  for  continuous 


174  CONTINUOUS  BEAMS  CHAP,  vill 

beams  under  uniform  loads»  It  was  first  deduced  by  Clapeyron 
in  1857  and  is  hence  sometimes  called  Clapeyron's  theorem. 

This  theorem  shows  how  the  moment  M"  at  any  support 
is  connected  with  moments  at  the  preceding  and  following 
supports.  When  all  spans  are  of  the  same  length  /  and  have 
the  same  load  w  per  linear  unit,  the  theorem  becomes, 

M'+4M"+M"'=  -%wl2  (69)' 

which  applies  to  the  most  common  cases  in  practice.  It  must 
be  noted,  however,  that  these  theorems  of  three  moments  are 
only  valid  when  the  beam  is  of  constant  section  area  and  when 
all  the  supports  are  on  the  same  level,  since  these  condition? 
have  been  introduced  into  the  algebraic  work  by  taking  /  as 
constant,  and  by  taking  y  as  zero  for  all  supports. 

^//////////////////////////y''     l".    M"f  ^^^^^^^^^^^ 

Fig.  69a  Fig.  696 

In  any  continuous  beam  of  s  spans  there  are  s  +  i  supports 
and  s  —  i  unknown  bending  moments  at  the  supports,,  For 
each  of  these  supports  an  equation  of  the  form  of  (69)  or  (69)' 
may  be  written  which  contains  three  unknown  moments.  Thus 
there  will  be  stated  s  —  i  equations,  and  the  solution  of  these 
will  furnish  the  values  of  the  5  —  1  unknown  quantities. 

The  simplest  case  is  that  shown  in  Fig.  69fr,  where  there 
are  two  equal  spans  uniformly  loaded,  the  left  and  right  ends 
of  the  beam  resting  upon  the  supports.  Here  M'  and  M'"  are 
each  zero,  and  the  theorem  (69)'  gives  Mn '=  —\wl2.  The  left 
reaction  R'  is  now  found  from  R!l  —  \wl1  =  M"  to  be  R'  =  $wl, 
and  R"f  has  the  same  value;  hence  each  span  of  Fig.  69£  is 
in  the  same  condition  as  that  of  a  beam  fixed  at  one  end  and 
supported  at  the  other. 

Prob.  69.  A  continuous  beam  of  two  spans  is  uniformly  loaded  with 
125  pounds  per  linear  foot.  The  length  of  the  first  span  is  18  feet  and 
that  of  the  second  span  is  1 2  feet.  Compute  the  moment  at  the  middle 
support,  and  the  three  reactions. 


ART.  70  EQUAL  SPANS  WITH  UNIFORM  LOAD  175 


ART.  70.     EQUAL  SPANS  WITH  UNIFORM  LOAD. 

Consider  a  continuous  beam  of  five  equal  spans  uniformly 
loaded.  Let  the  supports,  beginning  on  the  left,  be  numbered 
i,  2,  3,  4,  5,  and  6.  From  the  theorem  of  three  moments  an 
equation  may  be  written  for  each  of  the  supports  at  which 
moments  exist;  thus, 

foe  support  2  ,  M  i  +  4M2  +  M%  =  —  %wl2 

for  support  3  ,  M2  +  4^/3  +  M±=  —  %w  I2 

for  support  4  ,  M  3  +  4  M4  +  M5  =  —  %wl2 

for  support  5,  M±+4M$+M§=  —  %wl2 

Since  the  ends  of  the  beam  rest  on  abutments  without  restraint 
\fi  =  M6  =  o.  Hence  the  four  equations  furnish  the  means 
of  finding  the  four  moments  M2,  MS,  M^  M*,.  The  solution 
may  be  abridged  by  the  fact  that  M^  =  M^  and  Ms  =  M±,  which 
is  evident  from  the  symmetry  of  the  beam.  Hence, 


From  formula  (67)"  the  shears  at  the  right  of  the  supports  are, 

etc. 


From  (67)'  the  moment  for  any  section  in  any  span  may  now 
be  found  as  in  Art.  68,  and  by  the  methods  there  indicated  the 
complete  investigation  of  any  special  case  may  be  effected. 


Moments  at  Supports 
for  Equal  Spans. 


Fig.  70a 

In  this  way  the  moments  at  the  supports  for  any  number 
of  equal  spans  can  be  deduced.  The  following  triangular  table 
shows  their  values  for  spans  as  high  as  seven  in  number.  In 


176 


CONTINUOJS  BEAMS 


CHAP.  VIII 


each  horizontal  line  the  supports  are  represented  by  squares 
in  which  are  placed  the  coefficients  of  —  wl2.  For  example, 
in  a  beam  of  3  spans  there  are  four  supports,  and  the  bending 
moments  at  those  supports  are  o,  —  -^wl2,  —^wl2,  and  o. 

The  shears  at  the  supports  are  also  shown  in  the  following 
table  for  any  number  of  spans  less  than  six.  The  space  repre- 
senting a  support  gives  the  shear  on  the  left  of  the  support  in 
its  left-hand  division  and  the  shear  on  the  right  of  the  support 
in  its  right-hand  division.  The  sum  of  the  two  shears  for  any 
support  is,  of  course,  the  reaction  of  that  support.  For  example, 
in  a  beam  of  five  equal  spans  the  reaction  at  the  second  sup- 
port is 


Shears  at  Supports 
for  Equal  Spans. 


Fig.  706 

•  It  will  be  seen  on  examination  that  the  numbers  in  any  oblique 
column  of  these  tables  follow  a  certain  law  of  increase  by  which 
it  is  possible  to  extend  them,  if  desired,  to  a  greater  number  of 
spans  than  are  here  given. 

As  an  example,  let  it  be  required  to  select  a  rolled  steel  I  beam 
to  span  four  openings  of  8  feet  each,  the  load  per  span  being 
44  ooo  pounds  and  the  greatest  horizontal  stress  in  any  fiber  to  be 
15  ooo  pounds  per  square  inch.  The  required  beam  must  satisfy 
the  flexure  formula  S  .  I'/c  =  M,  or  it  must  be  of  such  size  that 
I/c  =  M/i$ooo.  From  the  table  it  is  seen  that  the  greatest 
negative  moment  is  that  at  the  second  support  or  -^wl2,  and 
the  maximum  positive  moment  in  the  first  span  is  Vi2/2W  = 
Tpgf-swP  and  that  in  the  second  span  is  M2  +  V22/2W  =  Tf  J¥7t> I2. 


ART.  71  UNEQUAL  SPANS  AND  LOADS  177 

The  greatest  value  of  M  is  hence  at  the  second  support;   then, 
7/^=3X44000X8X12/28X15  000=30.2  inches3 

and  from  Table  6  it  is  seen  that  the  light  1  2-inch  beam,  for 
which  I/c  is  36.0,  will  most  closely  satisfy  the  requirements. 

Prob.  70a.  Find  several  shears  and  moments  for  three  equal  spans 
uniformly  loaded,  and  draw  the  shear  and  moment  diagrams. 

Prob.  706.  Select  the  proper  steel  I  beam  to  span  three  openings  of 
12  feet  each,  the  uniform  load  on  each  span  being  6  ooo  pounds  and 
the  greatest  value  of  S  to  be  12  ooo  pounds  per  square  inch. 

ART.  71.    UNEQUAL  SPANS  AND  LOADS 

As  the  first  example,  consider  two  spans  with  lengths  /i,  l^ 
and  uniform  loads  per  linear  unit  w\  and  w2.  The  theorem  of 
three  moments  in  (69)  then  reduces  to, 

2M2(11  +  12)=~  M/!3  -  iw2/23 

from  which  the  bending  moment  at  the  middle  support  is  known. 
When  there  is  no  load  upon  the  second  span  w2  is  zero.  As  a 
particular  case  let  /i  =  4o  and  /2  =  3Q  feet,  Wi  =  2io  pounds  per 
linear  foot  and  w2  =  o;  then  M2=—  24000  pound-feet.  The 
reaction  RI  is  found  from  RI  X40  —  JX2ioX  4O2  =  M  2,  which  gives 
Rl=  +  3600  pounds.  The  reaction  RS  is  found  from  R3  X  30  =  M  2, 
which  gives  RS  =  —  800  pounds.  The  shear  passes  through 
zero  in  the  first  span  at  the  point  for  which  RI—  wx  =  o,  which 
gives  x  =  Ri/w,  and  the  maximum  positive  moment  is  then 
M  =  Ri2/2W  =  30  900  pound-feet.  From  these  values  the  shear 
and  moment  diagrams  in  Fig.  71  a  are  constructed. 

Next  consider  three  spans  having  the  lengths  /i,  /2,  and  /3, 
and  loaded  uniformly  with  Wi,  w2,  w3.  The  moments  at  the 
second  and  third  supports  are  M2  and  M%.  Then  from  (69), 


M212+2M3(12+13)=  - 

and  the  solution  of  these  gives  the  values  of  M2  and  M^.  A 
very  common  case  for  swing  drawbridges  is  that  where  two  end 
spans  are  equal  and  the  load  uniform  throughout,  or  12  =  /,  /i  =  /a= 


178  CONTINUOUS  BEAMS  CHAP,  vili 

a/,  and  w\  =  W2  =  Wz  =  w.     For  this  case  the  solution  gives, 


For  example  take  a  swing-bridge  where  the  two  end  spans  are 
each  120  feet  and  the  middle  span  is  24  feet,  this  being  a  con- 
tinuous girder  when  closed.  Here  0:=  120/24  =5,  and  M2  = 
—  2.423W/2,  which  is  the  moment  at  supports  2  and  3  due  to 
live  load  over  all  spans.  When  live  load  covers  only  the  first 
span  w2  may  be  made  zero,  and  the  moments  be  found  by  the 
solution  of  the  above  equations.  In  Part  IV  of  Roofs  and 
Bridges  these  cases  of  loading  are  fully  discussed. 


T*      TT"       ~?4 


^v^ 

Fig.  71a  Fig.  716 

Whatever  be  the  lengths  of  the  spans  or  the  intensity  of  the 
uniform  loads,  the  theorems  of  three  moments  in  Art.  69  furnish 
the  means  of  finding  the  bending  moments  at  the  supports. 
Then  by  the  methods  of  Art.  67  shears  and  moments  at  every 
section  may  be  computed  and  the  degree  of  security  of  the  beam 
be  investigated  by  the  flexure  formula  (41).  Finally,  if  the 
material  is  not  stressed  beyond  its  elastic  limit,  formula  (45) 
may  be  used  to  determine  the  deflection. 

Prob.  710.  A  continuous  beam  of  three  spans  is  loaded  only 
in  the  middle  span,  as  in  Fig.  716.  Find  the  reactions  of  the  end 
supports  due  to  this  load. 

Prob.  716.  A  heavy  1 2-inch  steel  I  beam  of  36  feet  length  covers 
four  openings,  the  two  end  ones  being  each  8  feet  and  the  others  each 
10  feet  in  span.  Find  the  maximum  moment  in  the  beam.  Then 
determine  the  load  per  linear  foot  so  that  the  greatest  horizontal  unit- 
stress  may  be  15  ooo  pounds  per  square  inch. 

Prob.  7  ic.  For  the  case  of  three  spans  let  the  first  and  third  spans 
be  each  80  feet  long.  Find  the  length  of  the  middle  span  so  that  the 
moment  shall  be  zero  at  the  middle  of  that  span,  the  load  being 
uniform  throughout. 


ART.  72  SPANS   WITH    FlXED    ENDS  179 

ART.  72.     SPANS  WITH  FIXED  ENDS 

The  theorem  of  three  moments  may  also  be  used  to  determine 
the  bending  moments  at  the  supports  when  the  ends  of  the  con- 
tinuous beam  are  horizontally  fixed  in  walls.  If  the  number 
of  spans  be  two,  there  are  three  unknown  moments  and  hence 
three  equations  are  to  be  written  for  four  spans;  in  these  the 
lengths  of  the  first  and  last  spans  are  to  be  made  zero  and  thus 
the  elastic  curve  will  be  made  horizontal  at  the  ends  of  the  beam. 
The  following  example  will  illustrate  the  procedure. 


[ 

£^     A  S  A     3 

12  3  45 

Fig.  726 

Let  there  be  two  spans  of  lengths  12  and  /3  with  ends  hori- 
zontally fixed,  as  in  Fig.  72a.  In  Fig.  72b  let  the  restraint  of 
the  walls  be  replaced  by  the  span  /i  on  the  left  end  and  the  span 
/4  on  the  right  end,  these  spans  being  taken  as  unloaded.  From 
(69)  the  equations  for  the  three  supports  2,  3,  4,  are, 

for  support  2  ,       Af  1/1  +  2M2  (/i  +  12)  +  M  3/2  =  —  1  W2123 

for  support  3,       Af2/2+2Af3(/2+/3)+Af4/3=  -  iw2/23  -  i^3/33 

for  support  4  ,       M313  -f  2  Af  4(/3  -f  /4)  +  M314  =  -  \w  3/33 

Now,  MI  and  M*>  are  zero  since  the  ends  are  supposed  to  merely 
rest  on  the  supports  i  and  5.  By  making  /i  =  o  the  points  i  and 
2  become  consecutive,  which  renders  the  elastic  curve  horizontal 
at  2  ;  also  by  mak'ng  /4  =  o  the  points  4  and  5  become  consecutive 
which  renders  the  elastic  curve  horizontal  at  4.  As  a  special 
case  let  I2  =  h  =  l  and  w2  =  w3  =  w,  so  that  the  two  spans  are  equal 
and  have  the  same  uniform  load  ;  then  from  symmetry  it  is  known 
that  Af4  is  equal  to  M2,  so  that  the  equations  become, 

2  M2  +  M3  =  -  $wl2  2  AT  2  +  4^3  =  -  i^2 

from  which  are  found  M2=  -T^wl2  and  M3=  —  -fawl2.  As 
another  special  case  let  the  two  spans  be  equal  and  the  uniform 
load  be  only  on  the  first  span  or  w2  =  w  and  w3  =  o;  then  the 
equations  are, 


180  CONTINUOUS  BEAMS  CHAP,  vni 

from  which  the  moment  at  the  left  fixed  end  is  M^=  —  ^wl2 
that  at  the  middle  support  is  M3=  —^wl2,  and  that  at  the  right 
fixed  end  is  M 4  =  +^wl2. 

The  fixing  of  the  ends  renders  the  bending  moments  smaller 
than  those  of  beams  with  supported  ends  and  hence  causes  an 
increase  of  strength,  while  the  stiffness  is  also  made  greater. 
Continuous  beams  in  the  floors  of  buildings  often  have  their 
ends  fixed  in  walls.  When  the  spans  are  two  in  number  and 
of  equal  length,  each  span  is  in  the  same  condition,  under  uni- 
form load,  as  one  with  both  ends  fixed,  since  the  elastic  curve 
is  horizontal  over  the  middle  support. 

Prob.  72<z.  Draw  the  shear  and  moment  diagrams  for  a  beam  of 
two  equal  spans  with  fixed  ends,  the  first  span  being  unloaded  and  the 
second  covered  with  uniform  load. 

Prob.  726.  Using  the  theorem  of  three  moments  for  concentrated 
loads  given  in  the  next  article,  deduce  the  moments  for  Fig.  72a  caused 
by  a  load  P  at  the  middle  of  the  first  span. 

ART.  73.     CONCENTRATED  LOADS 

Thus  far  only  uniform  loads  upon  one  or  more  spans  have 
been  discussed,  but  all  the  methods  given  are  applicable  to  con- 
centrated loads,  provided  the  moments  at  the  supports  due  to 
those  loads  can  be  found.  By  a  process  of  reasoning  similar 
to  that  in  Art.  69,  a  theorem  of  three  moments  for  such  loads 
can  be  deduced,  and  it  will  here  be  stated  without  the  algebraic 
work  of  demonstration.  As  before  the  beam  is  to  be  of  constant 
section  throughout  its  length  and  all  supports  are  to  be  upon 
the  same  level. 

Let  /'  and  I"  be  the  lengths  of  any  two  consecutive  spans  and 
M',  M",  and  M'"  the  moments  at  the  three  supports.  Let  P' 
be  any  load  upon  the  first  span  at  the  distance  td'  from  the  first 
support,  and  P"  any  load  upon  the  second  span  at  the  distance 
*/"  from  the  second  support,  K  being  any  fraction  less  than  unity 
and  not  necessarily  the  same  for  the  two  loads.  Then  the  theorem 
of  three  moments  is, 


ART.  73 


CONCENTRATED  LOADS 


181 


which  is  to  be  used  in  the  same  manner  as  those  of  Art.  69.  If 
there  is  no  load  on  the  span  /',  then  P'  is  zero;  if  there  is  none 
on  the  span  /",  then  P"  is  zero. 

To  illustrate  the  application  of  this  formula,  let  there  be 
two  equal  continuous  spans,  as  in  Fig.  73a,  with  a  load  P  on 
the  first  span.  Here  Pf  becomes  P  and  P"  is  zero,  and  since 
there  are  no  moments  at  the  ends,  the  theorem  gives  M^  = 
-JP/O-K3).  To  find  the  left  reaction,  Ril-P(l-Kl)  =  M2 
from  which  J?i  =  lP(4  — 5K+*3).  When  the  load  is  at  the  middle 
of  the  span,  K  is  J,  and  M%=  —-£%Pl  and  RI  =  ||P.  The  reaction 
at  the  right  end  is  found  from  R3l=M2)  whence  R$=  -\P(n- 
/c3)  or  R3=  —  33g-P;  hence  the  right  end  must  be  prevented  from 
rising  from  the  support  in  order  that  this  negative  reaction 
can  prevail.  If  that  end  is  not  fastened,  R3  is  zero  and  the 
reaction  RI  is  P(I  —  K)  as  for  a  simple  beam. 

U-- rf— Jp  ft  |P9 


Fig.  73a 


Fig.  735 


From  the  above  results  and  from  the  definitions  of  shear 
and  moment  in  Arts.  37  and  38,  the  shear  and  moment  dia- 
grams may  be  drawn,  as  in  Fig.  73a.  The  inflection  point  is 
found  from  Rioc  —  P(oc—Kl)  =  o,  whence  its  position  is  given  by 
#=4//(5-K2);  when  K=I  the  load  is  at  the  middle  support 
and  #=/;  when  K=O  the  load  is  at  the  left  support  and  x  =  %l. 
Hence  the  inflection  point  for  a  load  in  the  first  span  always 
lies  on  the  last  fifth  of  the  span, 

When  there  are  loads  on  both  spans,  as  in  Fig.  73  b,  the  moments 
due  to  both  may  be  found  from  the  theorem,  or  the  moments 
and  reactions  due  to  each  may  be  separately  determined  and 
the  final  moment  found  by  addition.  Thus,  if  each  load  is  at 
the  middle  of  the  span,  the  reaction  RI  due  to  P2  is  known  from 


182  CONTINUOUS  BEAMS  CHAP.  VIII 

the  value   of   the  reaction  R$  due  to  PI;   hence  for  both   loads 
Ri  =  HPi—fjP2,  so  that,  if  PI  and  P2  are  equal  PI  is  +  T5<rP. 

The  theorem  of  three  moments  above  given  was  deduced  in 
1865  by  Bresse,  and  from  it  the  theorem  of  Clapeyron  (Art.  69) 
for  uniform  loads  is  readily  derived.  Let  ix/  be  the  load  per 
linear  unit  on  the  first  span  and  P'  be  a  small  part  of  this  uni- 
form load  extending  over  the  distance  £(*/),  then  P'  is  to  be 
replaced  by  wd(id)  and  P'/'2(/c—  /c3)  becomes  w'l'3(K- K?)dK. 
Integrating  this  between  the  limits  i  and  o,  the  uniform  load 
covers  the  whole  span  and  the  function  is  \w'l'*  as  in  Clapeyron 's 
theorem. 

An  abbreviated  method  of  finding  the  moments  at  the  rap- 
ports, without  writing  the  theorem  of  three  moments  for  each 
support,  was  devised  by  the  author  in  1875.  See  London  Philo- 
sophical Magazine,  September,  1875;  or  Roofs  and  Bridges, 
Part  IV.  This  method  can  also  be  directly  applied  to  cases 
where  one  or  both  ends  are  fixed.  Continuous  beams  with  fixed 
ends  are,  however,  rarely  used  under  the  action  of  a  live  load. 

Prob.  73.  A  continuous  beam  has  four  spans  of  6,  8,  8,  6  feet 
length;  the  ends  resting  upon  abutments.  Find  the  left  reaction  due 
to  a  load  of  i  ooo  pounds  at  the  middle  of  the  second  span. 

ART.  74.     SUPPORTS  ON  DIFFERENT  LEVELS 

All  cases  of  flexure  thus  far  considered  have  been  for  sup- 
ports on  the  same  level,  except  that  of  fixed  beams  in  Art.  64. 
The  general  remarks  there  given  regarding  the  effect  of  changes 
of  level  of  the  supports  apply  also  to  continuous  beams.  Indeed 
a  slight  depression  of  one  support  below  the  level  of  the  others 
may  cause  great  changes  in  the  moments  and  stresses  through- 
out the  beam. 

Let  Fig.  74#  represent  two  consecutive  spans  of  a  continu- 
ous beam  having  the  lengths  I'  and  ln ';  let  the  axis  of  abscissas 
be  horizontal,  hf,  h",  and  h'n  being  the  heights  of  the  three  sup- 
ports above  this  axis.  Let  the  beam  be  anchored  to  the  sup- 
ports so  that  its  lower  surface  is  compelled  to  touch  them  under 
all  circumstances.  This  constraint  produces  moments  at  the 


ART.  74  SUPPORTS    ON    DIFFERENT    LEVELS  183 

supports,  the  magnitude  of  which  will  depend  upon  the  size 
and  shape  of  the  beam,  represented  by  the  moment  of  inertia  /, 
and  upon  the  stiffness  of  its  material,  represented  by  the  modulus 
of  elasticity  E. 

By  proceeding  as  in  Art.  69  a  theorem  of  three  moments 
for  this  case  may  be  deduced.  In  the  first  span,  y  =  hf  when 
x=o  and  y=h"  when  x  =  l'.  The  value  of  dy/dx  for  the  first 
and  second  spans  hence  differs  from  those  of  Art.  69  in  contain- 
ing the  quantities  h',  h",  h'" .  By  equating  the  values  of  dy/dx 
for  the  middle  support,  there  will  be  found, 


•6EI 


h"-hf    h"-h'"\ 
I'     +      I"      ) 


which  is  the  theorem  of  three  moments  for  uniform  load  and 
supports  on  different  levels.  This  may  be  extended  to  include 
concentrated  loads  by  inserting  the  functions  of  P  and  K  given 
in  the  last  article. 


Fig.  74<z  Fig.  746 

To  show  the  method  of  application  of  this  theorem,  take 
two  equal  continuous  spans  with  supported  ends,  and  let  the 
middle  support  be  lowered  the  distance  /  below  the  level  of  the 
other  supports.  Then  /'  =  I"  =  /  and  M'  =  M"r  =  o,  also  h"  -h'=* 
h"  —  h"'=  -/.  Let  the  load  be  uniform  throughout,  so  that 
•u/  =  lu/'  =  w.  Then  there  results, 


which  shows  that  the  negative  bending  moment  at  the  middle 
support  is  decreased  by  the  circumstance  of  its  depression.  When 
/  has  the  value  wl4/24EI,  there  is  no  moment  at  this  support 
and  each  span  is  like  a  simple  beam.  When  /  has  a  greater 
value,  the  moment  becomes  positive.  If  the  beam  be  one  of 
rolled  steel  weighing  40  pounds  per  foot  and  the  spans  be  16 
feet  long,  the  moment  at  the  support  due  to  the  weight  of  the 


184  CONTINUOUS  BEAMS  CHAP.  VIII 

beam  is  -Jw/2=  —  i  280  pound-feet.  Now  let  the  middle  sup- 
port be  depressed  o.i  inches  below  the  level  of  the  others;  then 
since  7=158.7  inches4  from  Table  6,  the  moment  due  to  that 
depression  is  $EIf/l2=  +  3  230  pound-feet,  so  that  this  slight 
depression  entirely  changes  the  character  of  the  stresses  through- 
out the  beam. 

Continuous  bridges  are  subject  to  all  the  uncertainties  of 
continuous  beams  in  regard  to  the  effect  of  changes  of  level  of 
the  supports,  and  hence  their  use  has  been  almost  entirely  aban- 
doned. Continuous  beams  are  used  only  for  short  spans  as  is 
the  case  with  railroad  rails,  and  in  floors  where  there  is  little 
liability  to  change  in  level  of  supports. 

In  conclusion  it  may  be  noted  that  the  above  theorem  of 
three  moments  furnishes  a  very  convenient  method  for  finding 
the  elastic  deflections  of  beams.  As  an  example,  take  the  case 
shown  in  Fig.  746,  where  it  is  desired  to  find  the  upward  deflec- 
tion at  the  middle  of  the  span  /  due  to  a  uniform  load  wm  on  the 
overhanging  end  m.  Let  the  middle  point  be  marked  2  and 
the  supports  i  and  3;  the  moments  at  these  points  are  M\  =  oy 
MZ*=  —  \wm2,  Mz  =  —  %wm2.  These  are  to  be  inserted  in  the 
formula  in  place  of  M' ',  M" ',  M'"\  also  /'  =  J/,  ln r  =  JJ,  and 
ivf  =  w"  =  o  since  there  is  no  load  on  the  span  considered.  Also 
making  h'  =  h"'  =  h,  the  deflection  is  h2  —  hi  and  the  formula 
now  gives  its  value  as  wm2l2/^2EI. 

Prob.  74#.  A  continuous  beam  of  two  equal  spans,  uniformly  loaded, 
has  its  supported  ends  on  the  same  level.  How  far  must  the  middle 
support  be  depressed  so  that  the  negative  moment  over  it  may  be 
numerically  equal  to  the  maximum  positive  moment  in  each  span  ? 

Prob.  746.  Find,  by  the  above  method,  upward  deflection  of,  the 
overhanging  end  in  Fig.  746,  due  to  a  uniform  load  over  the  span  /. 

ART.  75.    THE  THEORY  OF  FLEXURE 

The  theory  of  flexure,  presented  in  this  and  the  preceding 
chapters,  is  called  the  common  theory,  and  is  the  one  universally 
adopted  for  the  practical  investigation  of  beams.  It  should  not 
be  forgotten,  however^  that  the  axioms  and  laws  upon  which 


ART.  75  THE  THEORY  OF  FLEXURE  185 

it  is  founded  are  only  approximate  and  not  of  an  exact  nature 
like  those  of  mathematics.  The  law  regarding  the  proportionality 
of  stress  and  deformation  is,  for  instance,  only  roughly  approxi- 
mate for  brittle  materials.  The  flexure  formula  S  .  I/c  =  M  has 
been  established  from  this  law  and  from  the  observed  fact  that  a 
vertical  line,  drawn  upon  the  side  of  the  beam  before  flexure, 
remains  a  straight  line  after  flexure. 

When  the  load  on  a  beam  is  sufficient  to  cause  its  rupture, 
and  the  longitudinal  unit -stress  5  is  computed  from  the -flexure 
formula,  a  disagreement  of  that  value  with  those  found  by  direct 
experiments  on  tension  or  compression  is  observed.  This  is 
often  regarded  as  an  objection  to  the  common  theory  of  flexure, 
but  it  is  in  reality  no  objection,  since  the  laws  upon  which  the 
flexure  formula  is  founded  are  only  true  provided  the  elastic 
limit  of  the  material  is  not  exceeded.  Experiments  on  the  deflec- 
tion of  beams  furnish,  on  the  other  hand,  the  most  satisfactory 
confirmation  of  the  theory.  When  the  modulus  of  elasticity  E 
is  known  by  tensile  or  compressive  tests,  the  formulas  for  deflec- 
tion are  found  to  give  values  closely  agreeing  with  those  observed. 
Indeed  so  reliable  are  these  formulas  that  it  is  not  uncommon 
to  use  them  for  the  purpose  of  computing  E  from  experiments 
on  beams.  When,  however,  the  elastic  limit  of  the  material 
is  exceeded,  the  computed  and  observed  values  fail  to  agree. 

Certain  false  theories  of  flexure  have  been  proposed  from 
time  to  time,  the  one  best  known  being  that  in  which  it  is  assumed 
that  the  moment  of  the  horizontal  forces  on  one  side  of  the  neutral 
axis  is  equal  to  the  moment  of  those  on  the  other  side.  Since 
the  principles  of  static  equilibrium  furnish  no  condition  of  this 
kind,  the  formulas  established  are,  of  course,  without  value. 

Although  it  is  unfortunate  that  the  flexure  formula  does  not 
theoretically  apply  to  the  rupture  of  beams,  it  is  better  to  use 
it  for  such  cases  in  connection  with  experimental  constants 
(Art.  52)  than  to  employ  any  formula  which  disagrees  with  the 
fundamental  principles  of  statics.  Such  is  the  method  in  general 
practice,  and  on  the  whole  it  may  be  concluded  that  the  com- 
mon theory  of  flexure  is  entirely  satisfactory  and  that  it  is  suffi- 


186  CONTINUOUS  BEAMS  CHAP'VIH 

cient  for  the  investigation  of  most  questions  relating  to  the  strength 
and  stiffness  of  beams.  For  materials  like  cast-iron  and  con- 
crete it  is  possible  to  deduce  formulas,  which  apply  more  closely 
than  the  common  flexure  formula,  by  using  a  parabola  instead 
of  a  straight  line  to  represent  the  variation  of  the  stresses  above 
or  below  the  neutral  axis  (Art.  52).  These  formulas  include 
constants  which  give  the  relation  between  stress  and  deforma- 
tion, so  that  each  material  requires  a  different  flexure  formula. 
Although  such  formulas  are  theoretically  more  correct  than 
(41)  for  stresses  beyond  the  elastic  limit,  it  does  not  appear  that 
they  give  better  results  for  rupture  than  are  obtained  by  using 
(41)  with  the  values  of  Sf  found  by  experiment. 

In  all  the  examples  thus  far  discussed,  the  load  applied  to 
the  beam  is  parallel  to  one  of  the  principal  axes  of  inertia  of  the 
cross-section  and  its  resultant  coincides  with  that  axis.  When 
this  is  not  the  case,  the  flexure  formula  (41)  must  be  modi- 
fied in  the  manner  indicated  in  Art.  166;  such  unsymmetric 
arrangement  rarely  occurs  except  in  the  purlin  beams  of  roof 
trusses. 

The  actual  internal  stresses  in  beams  are  far  more  complex 
than  those  considered  in  the  common  theory,  because  the  ver- 
tical shears  combine  with  the  horizontal  stresses;  discussions 
of  the  apparent  and  true  stresses  are  given  in  Chapters  XI  and 
XV.  The  influence  of  shear  on  the  deflection  of  beams  is  investi- 
gated in  Art.  125.  All  the  formulas  and  methods  of  the  pre- 
ceding chapters  apply  only  to  beams  in  which  the  material  is 
the  same  throughout  the  section  area.  When  different  materials 
are  combined  to  form  a  beam,  the  flexure  formula  must  be  modi- 
fied so  as  to  take  into  account  their  different  degrees  of  stiffness, 
and  this  will  be  done  in  Chapter  XII. 

The  theory  of  beams  arose  from  the  discussions  of  Galileo  in 
the  seventeenth  century,  but  it  was  not  until  about  1825  that  the 
flexure  formula  and  the  general  equation  of  the  elastic  curve 
were  established  by  Navier.  Since  that  time  great  progress  has 
been  made  in  considering  the  flexure  of  beams  under  impact 
and  in  applying  the  principles  of  work  and  energy  to  their  dis- 


ART.  75  THE   THEORY    OF    FLEXURE  187 

cussion;  some  of  these  investigations  will  receive  attention  in 
future  chapters. 

Prob.  750.  A  beam  of  three  spans,  the  center  one  being  /  and  the 
side  ones  nl,  is  loaded  with  P  at  the  middle  of  each  span.  Find  the 
value  of  n  so  that  the  reactions  at  the  end  may  be  one-fourth  of  the 
other  reactions. 

Prob.  756.  Consult  Engineering  News,  vol.  xviii,  pp.  309,  352,  404, 
443;  vol.  xix,  pp,  n,  28,  48,  84;  and  vol.  xxii,  p.  131.  Write  an 
essay  concerning  certain  erroneous  views  regarding  the  theory  of  flex- 
ure which  are  there  discussed. 

Prob.  75c.  Procure  several  sticks  of  good  timber,  each  JXj  inches, 
and  of  lengths  about  8,  12,  and  16  inches.  Devise  and  conduct  experi- 
ments to  test  the  following  laws:  First,  the  strength  of  a  beam  varies 
directly  as  its  breadth  and  directly  as  the  square  of  its  depth.  Second, 
the  stiffness  of  a  beam  is  directly  as  its  breadth  and  directly  as  the  cube 
of  its  depth.  Third,  a  beam  fixed  at  the  ends  is  twice  as  strong  and 
four  times  as  stiff  as  a  simple  beam,  when  both  are  loaded  at  the 
middle. 


188  COLUMNS  OR  STRUTS  CHAP. 


CHAPTER  IX 
COLUMNS   OR  STRUTS 

ART.  76.     CROSS-SECTIONS  OF  COLUMNS 

When  a  prism  has  a  length  longer  than  about  eight  or  ten 
times  the  least  side  of  its  cross-section,  it  is  called  a  'column' 
or  'strut'.  When  the  length  of  the  prism  is  only  four  or  six 
times  as  long  as  the  least  side  of  its  cross-section,  the  case  is  one 
of  simple  compression  the  constants  for  which  are  given  in  Art.  5. 
Under  simple  compression  the  failure  occurs  for  brittle  mate- 
rials by  oblique  shearing  and  for  plastic  materials  by  enlarge- 
ment and  cracking  (Art.  18).  In  the  case  of  a  column,  how- 
ever, failure  is  apt  to  occur  by  a  sidewise  bending  which  causes 
flexural  stresses.  The  longer  the  column  the  greater  is  the  lia- 
bility to  lateral  flexure. 

Wooden  columns  are  usually  square  or  round,  and  when 
of  large  size  they  may  be  built  hollow.  Cast-iron  columns  are 
usually  round  and  hollow.  Wrought-iron  columns  were  built 
prior  to  1900  of  a  great  variety  of  forms,  but  structural  steel 
has  since  been  almost  entirely  used.  Rolled  I  beams  may  be 
used,  but  most  steel  columns  are  formed  by  riveting  togethei 
channels,  angles,  and  plates  (Art.  44).  Columns  are  exten- 
sively used  in  buildings  and  bridges.  A  piston-rod  of  a  steam- 
engine,  or  the  parallel  rod  of  a  locomotive,  is  a  column  when 
it  is  under  compression.  It  is  clear  that  a  square  or  round  sec- 
tion is  preferable  to  an  unsymmetrical  one,  since  then  the  lia- 
bility of  the  column  to  bend  is  the  same  in  all  directions.  For 
a  rectangular  section,  the  plane  of  flexure  will  evidently  be  per- 
pendicular to  the  longer  side  of  the  cross-section,  and  in  gen- 
eral the  plane  of  flexure  will  be  perpendicular  to  that  axis  of 
the  cross-section  for  which  the  moment  of  inertia  is  the  least; 
for  Art.  56  shows  that  the  deflection  of  a  beam  varies  inversely 
as  /,  In  designing  a  column  it  is  hence  advisable  that  the  cross- 


ART.  76 


CROSS-SECTIONS  OF  COLUMNS 


189 


section  should  be  so  arranged  that  the  moments  of  inertia  about 
the  two  principal  rectangular  axes  should  be  closely  equal. 

For  example,  let  it  be  required  to  construct  a  column  with 
two  I  beams,  as  in  Fig.  760,  the  pieces  connecting  the  flanges  being 
small  and  light  so  that  they  add  nothing  to  stiffness  or  strength. 
Let  the  beam  be  the  light  1 5-inch  size  weighing  42  pounds  per 
foot;  then  Table  6  gives  71  =  441.7  for  an  axis  perpendicular 
to  the  web  and  1 2=  14.62  inches4  for  an  axis  along  the  middle 
line  of  the  web;  also  the  section  area  a  =  12. 48  square  inches. 
Let  it  be  required  to  find  the  distance  oc  between  the  centers 
of  the  webs  so  that  the  moments  of  inertia  with  respect  to  axes 
through  the  center  of  gravity  of  the  column  section  shall  be 
equal.  For  the  axis  perpendicular  to  the  webs,  7=2X441.7; 
for  the  axis  parallel  to  the  webs,  7=  2X14.62  +  2X12. 48 (i#)2. 
Equating  these  two  values  and  solving  for  x  gives  #=11.70  inches. 


Fig.  76a  Fig.  766  Fig.  76c 

As  a  second  example,  take  the  section  in  Fig.  76&,  which  is 
formed  by  two  channels  and  two  plates,  the  rivets  being  omitted 
in  the  sketch.  Each  channel  is  10  inches  deep  and  weighs  35 
pounds  per  foot,  and  each  plate  is  x  inches  long  and  \  inch  thick. 
Using  Table  9,  the  moments  of  inertia  of  the  column  section 
with  respect  to  the  two  axes  through  its  center  of  gravity  are, 


1=  2[4.66+  10.29(^-0.70)2  +TV  Xo.5*3] 

Placing  these  expressions  equal,  the  value  of  x  is  found  to  be 
between  10  and  ioj  inches.  This  section  is  suitable  only  for  a 
column  less  than  6  feet  in  length,  as  the  riveting  of  the  plates 
to  the  angles  could  not  be  done  for  a  long  column. 

Fig.  76c  is  a.   section   frequently  used   for  bridge    members, 
there  being  but  one  plate  connecting  the  two  channels.     Here 


190  COLUMNS  OR  STRUTS  CHAP.  IX 

the  center  of  gravity  of  the  section  lies  above  a  line  drawn  through 
the  middle  of  the  webs  and  its  position  is  to  be  found  by  the 
method  of  Art.  42.  Then  by  the  principles  set  forth  in  Art.  43, 
the  moments  of  inertia  with  respect  to  the  two  rectangular  axes 
through  this  center  are  to  be  computed,  and  that  which  is  the 
smallest  is  to  be  used  in  the  column  formulas  given  in  the  follow- 
ing pages. 

Prob.  76<z.  Two  joists,  each  2X4  inches,  are  to  be  placed  6  inches 
apart  between  their  centers,  and  connected  by  two  others,  each  8  inches 
wide  and  x  inches  thick,  so  as  to  form  a  hollow  rectangular  column* 
Find  the  proper  value  of  x. 

Prob.  766.  Let  the  section  in  Fig.  76c  consist  of  a  plate,  f  X 12  inches, 
and  two  channels,  each  12  inches  deep  and  weighing  20  J  pounds  per 
linear  foot.  Compute  the  moments  of  inertia  with  respect  to  the  two 
axes  through  the  center  of  gravity. 

ART.  77.     DEFINITIONS  AND  PRINCIPLES 

When  a  short  prism  of  section  area  a  is  under  compression 
in  the  direction  of  its  length  and  the  resultant  force  P  acts  through 
the  centers  of  gravity  of  the  end  sections,  the  internal  stress  is 
uniformly  distributed  over  the  section,  and  hence  the  compres- 
sive  unit-stress  S  is  P/a.  For  a  long  prism,  or  column,  this  is  not 
always  the  case,  for  any  sidewise  deflection  will  cause  flexural  stress 
which  will  render  the  unit-stress  on  the  concave  side  of  the  column, 
greater  than  P/a  and  that  on  the  convex  side  less  than  P/a. 
Hence  for  any  given  section,  the  load  P  should  be  taken  smaller 
for  a  long  column  than  for  a  short  one,  since  evidently  the  liability 
to  bending  increases  with  the  length. 

The  'Axis'  of  a  column  is  the  line  passing  through  the  centers 
of  gravity  of  the  cross-sections.     When  the  column  is  straight, 
the  axis  is  a  straight  line;    if  it  bends  laterally,  the  axis  is  the 
elastic  curve.     An  'axial  load'  is  one  having  its  line  of  action 
coinciding  with  the  centers  of  gravity  of  the  two  end  sections; 
the  term  '  concentric  load '  is  used  by  some  writers  for  this  case 
The  load  P  is  regarded  as  axial  in  the  greater  part  of  this  chapter 
this  being  the  most  common  case  in  practice. 


ART.  77  DEFINITIONS   AND    PRINCIPLES  191 

The  length  of  a  column  is  indicated  by  /  and  the  least  radius 
of  gyration  of  its  cross-section  with  respect  to  an  axis  through 
the  center  of  gravity  of  that  section  by  r.  The  value  of  r  is  found 
from  the  equation  ar2  =  I  (Art.  43)  where  a  is  the  section  area 
and  /  is  the  least  moment  of  inertia;  for  example,  if  the  section 
is  a  circle  of  diameter  d,  the  value  of  a  is  \nd?  and  that  of  /  is 
ttxd4  (Art.  43) ;  hence  the  radius  of  gyration  of  a  circular  section 
is  \d.  For  a  rectangle  having  its  least  side  d  and  its  width  b, 
the  radius  of  gyration  is  found  from  r2  =  ^jbd3/bd=-fi2d2  whence 
r=  0.289^.  For  sections  of  rolled  beams  and  channels,  the  values 
of  r  for  two  rectangular  axes  are  given  in  Tables  6  and  9,  and 
the  least  of  these  is  the  one  needed  in  computations  when  a  beam 
or  channel  is  to  be  used  as  a  column. 

It  was  shown  in  Art.  51,  that  a  given  section  area  a  offers 
greater  resistance  to  flexure  the  further  the  material  is  removed 
from  the  neutral  axis.  When  a  column  bends  laterally,  flexural 
stresses  similar  to  those  in  a  beam  arise,  and  hence  for  columns 
economy  is  also  promoted  by  placing  the  material  as  far  as  prac- 
ticable from  the  axis  about  which  bending  may  occur;  this  is 
done  by  making  the  radius  of  gyration  as  large  as  practicable. 

The  ratio  l/r  is  called  the  '  slenderness  ratio  '  of  the  column. 
When  l/r  is  less  than  about  25,  the  column  is  a  short  prism  under 
simple  compression  (Art.  5) ;  when  l/r  is  greater  than  about  200 
the  column  is  called  long  and  failure  occurs  by  lateral  bending. 
The  columns  generally  used  in  engineering  practice  have  slender- 
ness  ratios  varying  from  50  to  r5o. 

The  condition  of  the  ends  of  columns  exerts  a  great  influence 
upon  their  strength.  'Round  ends'  are  those  which  are  free  to 
turn  upon  the  surfaces  where  they  abut;  Fig.  77 a  shows  one 
with  spherical  ends  and  also  one  where  the  compression  is  applied 
through  pins.  '  Fixed  ends '  are  those  subject  to  such  restraint 
that  the  tangent  to  the  elastic  curve  remains  vertical  at  the  ends 
when  a  lateral  deflection  occurs.  Fig.  77b  shows  a  column 
with  one  end  free  to  turn  and  the  other  fixed,  and  also  one  with 
both  ends  fixed.  Columns  with  both  ends  fixed  are  extensively 
used  in  buildings  and  bridges.  Columns  with  one  or  both  ends 


192 


COLUMNS  OR  STRUTS 


CHAP. 


hinged  on  pins  are  used  in  bridges  and  also  in  machines;  the 
piston-rod  of  a  steam-engine  is  a  case  of  a  column  with  one  end 
fixed  and  the  other  hinged.  The  term  c  round  ends '  generally 
includes  those  which  are  free  to  turn  on  pins  at  the  ends. 

It  is  evident  that  a  column  with  fixed  ends  is  stronger  than 
one  with  round  ends,  and  that  a  column  with  one  end  round  and 
the  other  fixed  is  intermediate  in  strength  between  these;  this 
is  confirmed  by  all  experiments.  There  is  also  another  condi- 
tion of  ends  which  is  called  'flat'  and  represented  in  Fig.  77c', 
here  the  ends  simply  abut  on  plane  surfaces  without  being  fixed. 
The  strength  of  a  column  with  flat  ends  is  closely  the  same  as 
one  with  fixed  ends  when  it  is  short,  and  about  the  same  as  one 
with  hinged  ends  when  it  is  long. 


Fig.  77a 


Fig.  776 


Fig.  77c 


In  the  following  articles  the  theory  of  columns  will  be  developed 
without  considering  the  weight  of  the  column  itself.  When  a 
column  is  in  a  vertical  position,  its  weight  brings  a  greater  unit- 
stress  upon  the  base  than  that  due  to  the  load  (Art.  27),  but  in 
most  practical  cases  this  increase  is  small.  When  a  column  is 
in  a  horizontal  position,  its  weight  causes  flexure  which  increases 
the  stress  on  the  upper  side  due  to  the  direct  compression,  and 
this  case  will  be  discussed  in  Arts.  101  and  102. 

Prob.  770.  A  round  cast-iron  column  has  the  outer  diameter  d\  and 
the  inner  diameter  d2.  Find  the  radius  of  gyration  of  the  cross-section. 

Prob.  77b.  An  I  beam  20  inches  deep  and  weighing  65  pounds  per 
linear  foot  is  used  as  a  column.  What  length  of  column  will  give  a 
slenderness  ratio  of  220? 


ART.  78.    EULER'S  FORMULA  FOR  LONG  COLUMNS 

Consider  a  long  column  of  section  area  a,  having  an  axial 
load  P  under  the  action  of  which  a  small  sidewise  bending  occurs. 


ART.  78  EuLER's    FORMULA  FOR   LONG    COLUMNS 


193 


The  column  is  supposed  to  be  perfectly  straight  before  the  appli- 
cation of  the  load,  but  in  practice  this  condition  is  not  attain- 
able, and  the  slightest  deviation  from  straightness,  or  a  lack  of 
homogeneity  in  the  material,  or  even  jars  and  shocks  due  to 
surrounding  objects,  causes  bending  to  occur  when  the  load 
P  is  sufficiently  large.  Even  for  a  perfectly  straight  column, 
it  is  found  when  P  reaches  a  certain  limit  and  a  slight  lateral 
force  is  applied  to  cause  lateral  flexure,  that  the  column  remains 
bent  when  this  lateral  force  is  removed.  It  is  required  to  find 
this  value  of  P. 

\f>  CL  b  C 

'"1 


Fig.  786 

Let  the  column  have  round  ends,  as  in  Fig.  78a,  and  let  / 
be  its  length,  I  the  least  moment  of  inertia  of  its  cross-section, 
and  E  the  modulus  of  elasticity  of  the  material.  Take  the  origin 
of  coordinates  at  the  upper  end,  and  let  x  be  measured  down- 
wards and  y  horizontal.  The  general  equation  (45)  is  regarded 
as  applicable  to  all  bars  subject  to  flexure,  provided  the  bending 
is  slight  and  the  elastic  limit  of  the  material  is  not  exceeded. 
For  the  case  in  hand  the  bending  moment  is  M  =  —  Py,  hence 
El .  d2y/dx2  =  —Py,  and  the  integration  of  this  gives, 


To  determine  C,  let  /  be  the  maximum  deflection  at  the  middle 
of  the  column  and  note  that  the  tangent  dy/dx  =  o  when  y=f\ 
hence  C  =  Pf2,  and  the  equation  may  be  written, 


Integrating  this,   and  determining  the   constant  by  the   condi- 
tion that  y  =  o  when  x  =  o,  the  equation  of  the  elastic  curve  is, 


194  COLUMNS  OR  STRUTS  CHAP.  IX 


x=  (EI/P)*  arc  sin(y/f  )  or  y=f  si 

The  curve  represented  by  this  equation  must  also  fulfill  the 
requirement  that  y  =  o  when  x  =  l.  This  condition  is  satisfied 
by  making  (P/EI)*l  equal  to  TT  or  some  integral  multiple  thereof; 
for,  if  v  is  any  integer,  then  sinvTr  =  o.  Accordingly, 

(P/E/)*/=i*r  or  P=y27T2E///2  (78) 

which  is  Euler's  formula  for  long  columns.  When  P  has  this 
value,  the  long  column  remains  bent  at  any  deflection  /  which 
may  happen  to  occur.  There  is  no  way  of  finding  /  from  the 
above  investigation  since  it  cancels  out  of  the  equation;  that 
is,  the  deflection  is  indeterminate. 

Inserting  the  value  just  found  for  (P/EI)*  in  the  equation 
of  the  elastic  curve,  it  reduces  to  the  form, 

x=(l/w)  arc  sin  (y/f)  or  y=fsinvx(x/I) 

By  discussing  this  equation  according  to  the  methods  of  An- 
alytic Geometry  there  are  derived  three  curves  for  v=i,  v  =  2, 
and  v  =  3,  as  shown  in  Fig.  786.  For  y  =  i  the  curve  is  entirely 
on  one  side  of  the  axis  of  x\  for  v  =  2,  it  crosses  that  axis  at  the 
middle;  for  v  =  3,  it  crosses  at  \l  and  §/.  Each  of  these  cases 
is  liable  to  occur  for  a  column  with  round  ends,  but  the  first 
is  the  most  dangerous  case  since  the  lateral  deflection  is  then 
the  greatest.  Hence,  making  v=i  in  (78),  there  is  found, 

P=7T2E///2  or  P/a=n2E(r/l)2 

which  is  Euler's  formula  for  columns  with  round  ends.  The 
second  form  is  obtained  from  the  first  by  using  I  =  ar2,  where  r 
is  the  least  radius  of  gyration  of  the  cross-section  (Art.  77). 

A  column  with  one  end  fixed  and  the  other  round  is  approxi- 
mately represented  by  the  part  b'b"  of  the  second  diagram  in 
Fig.  786,  where  V  is  the  fixed  end  where  the  tangent  to  the  curve 
is  vertical.  Here  v=2  and  the  length  b'b"  is  three-fourths  of  the 
entire  length;  hence  replacing  /  by  f/  in  (78),  it  becomes 
P=2{n2EI/l2  which  is  Euler's  formula  as  commonly  stated  for 
columns  having  one  end  fixed  and  the  other  round.  The  con- 
stant 2\  is,  however,  obtained  under  the  false  supposition  that 


ART.  78  EULER'S   FORMULA  FOR  LONG   COLUMNS  195 

the  point  b'  is  in  the  line  of  application  of  the  load  P.  The  more 
correct  analysis  in  Art.  88  shows,  however,  that  the  value  of  the 
constant  is  2.0457,  a  sufficiently  close  value  for  all  common  dis- 
cussions being  2.05.  Accordingly, 

P=2.o$7t2EI/P  or  P/a=2.o57:2E(r/l)2 

is  Euler's  formula  for  long  columns  having  one  end  fixed  and  the 
other  round. 

A  column  with  fixed  ends  is  represented  by  the  portion  cfc" 
of  the  third  case.  Here  v  =  3,  and  the  length  c'c"  is  two-thirds 
of  the  entire  length;  hence,  replacing  /  in  (78)  by  f/, 

P=4x2EI/l2  or  P/a=4x2E(r/l)2 

which  is  Euler's  formula  for  long  columns  with  fixed  ends. 

From  this  investigation  it  appears  that  the  relative  strengths 
of  long  columns  of  the  three  classes  are  as  the  numbers  i,  2.05, 
and  4  when  the  lengths  are  the  same,  and  this  conclusion  is 
approximately  verified  by  experiments.  A  general  expression 
for  Euler's  formula  for  long  columns  may  now  be  written,  namely, 

P=fiEI/l2  or  P/a=  fiE(r/l)2  (78)' 

in  which  the  number  /<  is  7r2  for  round  ends,  4;r2  for  fixed  ends, 
and  2.057T2  for  one  end  round  and  the  other  end  fixed. 

Another  kind  of  column  is  that  which  is  fixed  at  one  end 
and  entirely  free  at  the  other,  like  a  vertical  post  planted  in  the 
ground.  This  case  is  represented  in  Fig.  786  by  the  upper  half 
of  the  case  for  which  v=i,  by  the  upper  fourth  of  the  case  for 
which  y  =  2,  and  by  the  upper  sixth  of  the  case  for  which  1^  =  3. 
Using  either  case,  and  letting  /  be  the  length  of  the  column  under 
consideration,  there  is  found, 

P=ln2EI/l2  or  P/a=fr2E(r/l)2 

and  hence  the  number  /*  in  (78)'  is  \n2  for  a  long  column  fixed 
at  one  end  and  entirely  free  at  the  other.  Accordingly  a  column 
of  this  kind  can  carry  only  one-fourth  of  the  load  of  a  column 
with  two  round  ends. 

The  value  of  P  in  Euler's  formula  gives  the  axial  load  which 
holds  the  column  in  equilibrium  when  it  has  become  laterally 


196  COLUMNS  OR  STRUTS         CHAP.  IX 

deflected.  If  the  load  is  less  than  this  value  of  P,  the  column 
will  return  to  its  original  straight  position.  If  the  load  is  slightly 
greater  than  P,  the  bending  increases  until  failure  occurs. 
Euler's  formula  is  hence  the  criterion  of  indifferent  equilibrium, 
or  the  condition  for  the  failure  of  a  column  by  lateral  flexure. 

Euler's  formula  is  but  little  used  in  the  design  of  columns, 
except  in  Germany.  When  so  used  the  value  of  P  computed 
from  the  formula  is  to  be  divided  by  a  factor  of  safety  in  order 
to  give  the  safe  load  on  the  column. 

Prob.  7Sa.  A  solid  steel  column  with  round  ends  is  6  inches  in 
diameter  and  37  feet  long.  Compute  the  axial  load  which  will  cause 
it  to  fail  by  lateral  flexure. 

Prob.  786.  A  square  wooden  column  with  fixed  ends  is  20  feet  long 
and  carries  a  load  of  9  500  pounds.  Compute  its  size  so  that  it  may 
have  a  factor  of  safety  of  10  by  Euler's  formula. 

ART.  79.     EXPERIMENTS  ON  COLUMNS 

Although  Euler  published  his  formula  in  1757  and  Lagrange 
gave  a  more  satisfactory  discussion  of  it  in  1773,  it  was  not  until 
after  1825  that  its  conclusions  began  to  be  used  in  practical 
investigations.  This  formula  shows  that  the  load  P  which 
causes  the  failure  of  a  long  column  is  inversely  proportional  to 
the  square  of  its  length.  Hodgkinson  in  his  experiments  made 
about  1840  observed  that  this  was  closely  true  for  wrought-iron 
columns,  and  only  approximately  so  for  cast-iron  ones.  Since 
for  a  solid  cylindrical  column  I^-^xd4,  the  load  P  should  be 
proportional  to  the  fourth  power  of  the  diameter,  and  Hodgkin- 
son observed  that  this  ratio  was  a  little  too  high.  He  accordingly 
wrote  for  each  kind  of  columns  the  analogous  formula  P  =  Q  .  da/tf 
and  determined  the  constants  Q,  a,  /?  from  the  results  of  his 
experiments,  thus  producing  empirical  formulas. 

Let  P  be  the  load  in  gross  tons  which  causes  failure,  d  the 
diameter  of  the  column  in  inches,  and  /  its  length  in  feet.  Then 
the  empirical  formulas  deduced  by  Hodgkinson  for  solid  cylin- 
drical columns  are, 


ART.  79  EXPERIMENTS    ON   COLUMNS  197 

Cast  Iron  i  P=  M^3'5/^3  for  round  ends 

JP^^3'5/^63  for  flat  ends 

P=42d3'76l2  for  round  ends 


w™  ,aht  T  J      =42'l     or  roun    en 

Wrought  Iron      |  p^^pfyp  for  flat  ends 

These  formulas  indicate  that  the  ultimate  strength  of  flat-ended 
columns  is  about  three  times  that  of  round  -ended  ones.  The 
experiments  also  showed  that  the  strength  of  a  column  with 
one  end  flat  and  the  other  end  round  is  about  twice  that  of  one 
having  both  ends  round.  Hodgkinson's  tests  were  made  upon 
small  columns  and  his  formulas  are  not  so  reliable  as  those  which 
will  be  given  in  the  following  articles.  For  small  cast-iron 
columns  however  the  formulas  are  still  valuable.  A  flat  end 
may  sometimes  have  more  or  less  motion  when  the  deflection 
begins  and  hence  a  flat-ended  long  column  is  not  as  strong  as 
one  with  fixed  ends. 

After  1850  wrought  iron  slowly  replaced  cast  iron  as  a  struc- 
tural material,  and  many  tests  of  wrought-iron  columns  were 
conducted  prior  to  1890.  The  series  of  tests  made  by  Christie 
in  1883  for  the  Pencoyd  Iron  Works  is  of  great  value  on  account 
of  completeness  as  regards  wrought-iron  struts,  since  it  included 
angle,  tee,  beam,  and  channel  sections.  A  brief  description 
and  the  principal  results  will  here  be  given,  but  a  fuller  account 
may  be  found  in  Transactions  of  the  American  Society  of  Civil 
Engineers,  April,  1884. 

The  ends  of  the  struts  were  arranged  in  different  methods: 
first  flat  ends  between  parallel  plates  to  which  the  specimen  was 
in  no  way  connected;  second,  fixed  ends,  or  ends  rigidly  clamped; 
third,  hinged  ends,  or  ends  fitted  to  hemispherical  balls  and 
sockets  or  cylindrical  pins;  fourth,  round  ends,  or  ends  fitted 
to  balls  resting  on  flat  plates. 

The  number  of  experiments  was  about  three  hundred,  of 
which  about  one-third  were  upon  angles,  and  one-third  upon 
tees.  The  quality  of  the  wrought  iron  was  about  as  follows: 
elastic  limit  32  ooo  pounds  per  square  inch,  ultimate  tensile 
strength  49  600  pounds  per  square  inch,  ultimate  elongation 
1  8  percent  in  8  inches.  The  length  of  the  specimens  varied 
from  6  inches  up  to  16  feet,  and  the  ratio  of  length  to  least  radius 


198 


COLUMNS  OR  STRUTS 


CHAP.  IX 


of  gyration  varied  from  20  to  480.  Each  specimen  was  placed 
in  a  Fairbanks  testing  machine  of  50000  pounds  capacity  and 
the  power  applied  by  hand  through  a  system  of  gearing  to  two 
rigidly  parallel  plates  between  which  the  specimen  was  placed 
in  a  vertical  position.  The  pressure  or  load  was  measured  on 
an  ordinary  scale  beam,  pivoted  on  knife-edges  and  carrying  a 
moving  weight  which  registered  the  pressure  automatically.  At 
each  increment  of  5  ooo  pounds,  the  lateral  deflection  of  the  column 
was  measured.  The  load  was  increased  until  failure  occurred. 

The  following  are  the  combined  average  results  of  these  care- 
fully conducted  experiments.     The  first  column  gives  the  values 


Ratio  l/r  of 
Length  to 
Least  Radius 
of  Gyration 

Ultimate  Load  P/a,  in  Pounds  per  Square  Inch 

Fixed  Ends 

Flat  Ends 

Hinged  Ends 

Round  Ends 

20 

46  ooo 

46  ooo 

46  ooo 

44  ooo 

40 

40  ooo 

40  ooo 

40  ooo 

36500 

60 

36000 

36000 

36  ooo 

30500 

80 

32  ooo 

32  ooo 

31  5°° 

25  ooo 

100 

30000 

29  800 

28  ooo 

20  500 

120  • 

28  ooo 

26  300 

24  300 

1  6  500 

140 

25  500 

23500 

21  000 

12  800 

160 

23000 

20  000 

1  6  500 

9  500 

1  80 

20  OOO 

16800 

12  800 

7  5°° 

2OO 

17500 

14500 

10  800 

6  ooo 

220 

15  ooo 

12  70O 

8800 

5  00° 

240 

13000 

II  200 

7500 

4300 

260 

II  000 

9  800 

6  500 

3800 

280 

IO  OOO 

8500 

5  700 

3  200 

300 

9  ooo 

7  200 

5000 

2  800 

320 

8000 

6  ooo 

4  500 

2  500 

340 

7  ooo 

5  ioo 

4  ooo 

2  IOO 

360 

6  500 

4300 

35oo 

I  900 

380 

5  800 

3  5oo 

3000 

I  700 

400 

5  200 

3000 

2  5OO 

I  500 

420 

4800 

2  500 

2  300 

I  300 

440 

4300 

2  20O 

2  100 

460 

3800 

2  000 

I  90O 

480 

I  900 

I  800 

of  the  ratio  l/r  and  the  other  columns  the  values  of  P/a  which 
caused  failure,  these  being  the  ultimate  load  in  pounds  per  square 


ART.  79 


EXPERIMENTS  ON  COLUMNS 


199 


inch.  From  the  results  it  will  be  seen  that  there  is  little  prac- 
tical difference  between  the  strength  of  the  four  classes  when  the 
strut  is  short.  The  strength  of  the  long  columns  with  round 
ends  appears  to  be  about  one-third  that  of  those  with  fixed  ends. 
For  values  of  l/r  greater  than  200,  the  ultimate  loads  are  closely 
inversely  proportional  to  the  squares  of  the  lengths  for  round 
ends,  and  approximately  so  for  other  arrangements  of  ends. 

Euler's  formula  fairly  represents  the  results  of  the  tests  on 
the  long  columns.  Taking  £  =  25000000  pounds  per  square 
inch  and  n2  as  10,  the  formula  for  round-ended  columns  becomes, 

P/a=  250  ooo  ooo(r//)2=  250  ooo  ooc/(//r)2 
from  which  the  ultimate  unit-loads  are  computed, 

for    l/r=   220,          260,         300,         340,         380,         420 
P/a=$2oo,        3700,        2800,        2200,        1700,        1400 

while  the  experiments  give  the  ultimate  unit-loads  as, 

P/a=5ooo,       3800,        2800,        2100,        1700,        1300 
Since  Euler's  formula  is  deduced  under  the  laws  of  elasticity, 
it  must  be  concluded  that  the  elastic  limit  was  not  exceeded 
when  these  long  columns  failed  by  lateral  flexure. 


100  200  300 

Values  of  the  ratio  l/r 


400 


Fig.  79 

Fig.  79  gives  graphic  representations  of  the  above  results 
for  the  cases  of  fixed  ends  and  round  ends,  the  values  of  l/r  being 
taken  as  abscissas  and  those  of  P/a  as  ordinates.  The  broken 
lines  also  show  the  curves  for  these  two  cases  which  have  been 


200  COLUMNS  OR  STRUTS         CHAP,  ix 

plotted  from  Euler's  formulas.  It  is  seen  that  there  is  a  marked 
disagreement  between  the  experimental  results  and  those  found 
from  Euler's  formula  when  l/r  is  less  than  200;  this  disagree- 
ment is  due  to  the  circumstance  that  Euler's  formula  refers  to 
failure  by  lateral  bending  only,  while  for  values  less  than  200 
or  150  the  failure  actually  occurred  through  the  unit-stress  on 
the  concave  side  of  the  column  having  exceeded  the  elastic  limit, 
so  that  the  wrought  iron  became  plastic  (Art.  18). 

Prob.  79a.  A  cast-iron  cylindrical  column  with  flat  ends  is  to  be  7 
feet  long  and  carry  a  load  of  200  ooo  pounds  with  a  factor  of  safety  of 
6.  Compute  the  proper  diameter. 

Prob.  79b.  Let  Euler's  formula  be  written  y=c/x2J  where  x  and  y 
represent  l/r  and  P/a  and  c  is  a  constant.  Discuss  this  curve  and 
ascertain  the  points  where  it  is  parallel  to  the  coordinate  axes. 

ART.  80.     RANKINE'S  FORMULA 

The  columns  generally  employed  in  engineering  practice 
are  intermediate  in  length  between  short  prisms  and  the  long 
columns  to  which  Euler's  formula  applies.  They  fail  under 
the  stresses  caused  by  combined  flexure  and  compression,  columns 
of  brittle  material  by  oblique  shearing  on  the  concave  side  or 
by  tension  on  the  convex  side,  and  those  of  wrought  iron  and 
steel  by  the  flow  of  metal  on  the  concave  side  after  the  elastic 
limit  has  been  surpassed.  The  ultimate  unit-load  P/a  for  these 
columns  is  less  than  the  compressive  strength  Sc  for  short  prisms 
and  very  much  less  than  the  values  computed  from  Euler's 
formula,  as  Fig.  79  shows. 

When  such  a  column  is  perfectly  straight,  an  axial  load  P 
produces  the  same  unit-stress  S  =  P/a  on  all  parts  of  every  sec- 
tion area  a.  When  any  bending  occurs,  due  to  imperfections 
of  the  material  or  to  lack  of  straightness,  the  unit-stress  on  the 
concave  side  becomes  greater  than  P/a  and  that  on  the  convex 
side  becomes  less.  Fig.  80a  shows  the  flexure  very  much  exagger- 
ated; and  it  is  clear  that  the  flexure  formula  (41)  will  apply  to 
the  discussion  of  the  stresses  caused  by  lateral  bending.  Let  S\ 
be  the  greatest  unit-stress  due  to  the  flexure  and  P/a  the  average 


ART.  80 


RANKINE'S  FORMULA 


201 


unit-stress  due  to  the  direct  compression;  then  the  total  unit- 
stress  on  the  concave  side  is  the  sum  of  P/a  and  Si,  and  failure 
may  be  considered  as  occurring  when  this  sum  is  equal  to  the 
ultimate  compressive  strength  of  the  material. 


Fig.  80a 


Fig.  80c 


Let  /  be  the  length  of  the  column,  a  its  section  area,  /  the 
least  moment  of  inertia,  r  the  least  radius  of  gyration  of  that 
section,  and  c  the  distance  from  the  axis  of  the  column  to  the 
remotest  fiber  on  the  concave  side.  In  Fig.  80&,  the  average 
compressive  unit-stress  on  any  section  is  represented  by  cd,  but 
on  the  concave  side  this  is  increased  to  as  and  on  the  convex 
side  decreased  to  U.  The  triangles  pds  and  qdt  represent  the 
effect  of  the  flexure  exactly  as  in  the  case  of  beams,  ps  indicat- 
ing the  greatest  compressive  and  qt  the  greatest  tensile  unit- 
stress  due  to  the  bending.  Let  the  total  maximum  unit-stress 
as  be  denoted  by  S  and  the  part  due  to  the  flexure  be  denoted 
by  Si.  Then  51  =  P/a + Si,  in  which  Si  is  to  be  expressed  in  terms 
of  P  from  the  flexure  formula  S\  .  I/c  =  M,  where  M  is  the  bend- 
ing moment  due  to  P.  Let  /  be  the  maximum  lateral  deflection 
of  the  column;  then  the  greatest  value  of  M  is  Pf,  and  accord- 
ingly Si  =  Pcf/I,  or  replacing  /  by  ar2,  the  flexural  unit-stress 
Si  =  P/a .  cf/r2.  Accordingly,  the  greatest  compressive  unit- 
stress  on  the  concave  side  of  the  column  is, 


a     a 


By  analogy  with  the  theory  of  beams,  as  in  Art.  56,  the  deflection/ 
may  be  regarded  as  varying  directly  as  l2/c.  Hence,  if  0  is  a 
number  depending  upon  the  kind  of  material  and  the  condition 


202 


COLUMNS  OR  STRUTS 


CHAP.  IX 


of  the  ends  of  the  column,  it  follows  that, 

or 


(80) 


which  is  Rankine's  formula  for  the  investigation  of  columns. 


The  above  reasoning  has  been  without  reference  to  the  arrange- 
ment of  the  ends  of  the  column.  By  Art.  78  it  is  known  that 
a  long  column  with  fixed  ends  is  four  times  as  strong  as  one  with 
round  ends,  and  that  a  long  column  with  one  end  fixed  and  the 
other  end  round  is  2.05  times  as  strong  as  one  with  round  ends. 
Therefore,  assuming  that  similar  laws  hold  with  respect  to  the  term 
<j>(l/r)2  in  the  above  formula,  let  </>i  be  the  constant  for  fixed  ends, 
then  the  constant  for  round  ends  is  4^1  and  the  constant  for  one 
end  fixed  and  the  other  end  round  is  1.95^1.  The  remarks  near  the 
end  of  the  last  paragraph  of  this  article  verify  this  assumption. 

Values  of  5  and  <j>  may  be  determined  by  making  two  experi- 
ments on  columns  having  different  values  of  (l/r)  and  increasing 
the  loads  until  rupture  occurs,  thus  obtaining  different  values 
of  P/a.  From  (80)  two  equations  may  be  written  containing 
the  experimental  results  and  the  two  unknown  quantities  S  and  </), 
the  values  of  which  are  then  found  by  solution.  Wide  varia- 
tions are  found  in  the  results  thus  obtained  for  0  from  experi- 
ments on  the  rupture  of  different  types  of  columns,  but  the  fol- 
lowing table  gives  average  values  which  are  extensively  employed 
in  engineering  practice : 

VALUES  OF  <£  FOR  FORMULA  (80) 


Material 

Both  Ends 
Fixed 

Fixed  and 
Round 

Both  Ends 
Round 

Timber 
Cast  Iron 
Wrought  Iron 
Steel 

I 

1-95 

4 

3000 

I 

3  ooo 
i-95 

3000 
4 

5  °°° 

i 

5  °°° 
i-QS 

5  ooo 

4 
36000 

4 
25  ooo 

36000 

i 

36  ooo 
i-95 

25  ooo 

25  ooo 

AKT.  81  INVESTIGATION  OF  COLUMNS  203 

These  values  of  <f>  will  be  used  in  the  examples  and  problems  of 
the  three  following  articles,  and  the  value  to  be  taken  for  5  will 
be  ultimate  compressive  strength  of  the  material  for  cases  of 
rupture  and  the  allowable  compressive  unit-stress  for  cases  of 
design. 

Euler's  formula  is  not  satisfactory  for  practical  investigations 
because  it  contains  no  constant  indicating  the  .ultimate  or  work- 
ing strength  of  the  material  and  because  it  applies  only  to  long 
columns  for  which  the  ratio  l/r  is  greater  than  about  200. 
Rankinc's  formula,  however,  contains  the  constant  S  and  applies 
to  cases  for  which  the  ratio  l/r  lies  between  20  and  about  150, 
and  these  are  the  columns  used  in  engineering  practice.  On 
account  of  the  many  assumptions  employed  in  deducing  it,  and 
because  the  values  of  the  number  <J>  are  derived  from  experiments, 
the  formula  is  empirical  rather  than  rational,  yet  it  is  of  very 
great  value.  Its  form  satisfies  the  limiting  conditions  for  short 
and  long  prisms.  For  a  short  prism,  l/r  may  be  taken  as  zero, 
and  then  S-^P/a.  For  a  long  column,  unity  may  be  neglected 
in  comparison  with  <£(//r)2,  and  then  P/a  =  Sr2/l2(f>')  this  is  the 
same  in  form  as  Euler's  formula,  for  placing  S/(j>  equal  to  a  con- 
stant C  it  becomes  P  =  Ca.r2/l2.  Rankine's  formula  is  some- 
times referred  to  as  Gordon's  formula,  but  Gordon  used  the  least 
thickness  of  the  column  instead  of  the  least  radius  of  gyration. 

Prob.  80a.  Taking  values  of  l/r  as  abscissas  and  those  of  P/a  as 
ordinates,  discuss  the  curve  of  formula  (80)  and  find  where  its  tan- 
gents are  horizontal.  Also  locate  its  inflection  point. 

Prob.  SGb.  Plot  the  curve  represented  by  formula  (80)  for  wrought- 
iron  columns  with  fixed  ends,  taking  values  of  l/r  as  abscissas  and 
those  of  P/a  as  ordinates,  and  using  5  as  46  500  pounds  per  square 
inch.  Compare  the  plot  with  Fig.  79. 

ART.  81.    INVESTIGATION  OF  COLUMNS 

The  investigation  of  a  column  consists  in  determining  the 
maximum  compressive  unit-stress  S  from  formula  (80).  The 
values  of  P,  0,  /,  and  r  are  known  from  the  data  of  the 
given  case,  and  <j>  is  known  from  the  average  experimental 


204  COLUMNS  OR  STRUTS  CHAP.  IX 

values  given  in  the  table  of  the  last  article.      Then  the  value 
of  the  greatest  unit-stress  S  is  computed  from, 


By  comparing  the  computed  value  of  5  with  the  ultimate  com- 
pressive  strength  and  elastic  limit  of  the  material,  the  factor  of 
safety  and  the  degree  of  stability  of  the  column  may  be  inferred. 

For  example,  consider  a  hollow  wooden  column  of  rectangu- 
lar section,  the  outside  dimensions  being  4X5  inches  and  the 
inside  dimensions  3X4  inches.  Let  the  length  be  18  feet,  the 
ends  fixed,  and  the  load  be  5  400  pounds.  Here  P  =  5  400, 
a  =  8  square  inches,  /  =  2  16  inches,  and  <£  =  Tinnr-  The  least  radius 
of  gyration  is  that  with  respect  to  an  axis  parallel  to  the  longer 
side  of  the  section,  and  for  this  axis  r2  =  TV(5  X43  —  4X33)/8  =  2.21, 
and  accordingly  7/^=145.  Substituting  now  all  values  in  the 
formula,  there  is  found  5  =  5430  pounds  per  square  inch,  so 
that  the  factor  of  safety  is  only  about  ij.  The  average  unit- 
stress  for  this  case  is  the  same  as  for  a  short  prism  or  S  =  P/a  = 
675  pounds  per  square  inch,  which  is  a  safe  value  since  the  factor 
of  safety  is  nearly  12.  If  the  column  is  3  feet  long,  the  ratio  of 
slenderness  is  //r=24,  and  the  formula  gives  5  =  805  pounds 
per  square  inch,  which  corresponds  to  a  factor  of  safety  of  10. 

As  another  example,  consider  a  steel  column  21  feet  long 
with  fixed  ends  which  is  used  in  the  upper  chord  of  a  bridge 
under  an  axial  compression  of  240  ooo  pounds.  Let  the  sec- 
tion be  that  in  Fig.  76c,  which  consists  of  a  plate  f  Xi6  inches, 
and  two  channels  each  12  inches  deep  and  weighing  20^  pounds 
per  linear  foot.  From  the  principles  of  Arts.  42  and  43,  with 
the  help  of  Table  9,  the  moment  of  inertia  of  the  section  with 
respect  to  an  axis  through  its  center  of  gravity  and  perpen- 
dicular to  the  webs  is  found  to  be  501.4  inches4  and  that  with 
respect  to  an  axis  through  the  center  of  gravity  and  parallel 
to  the  webs  is  663.9  inches4.  The  least  radius  of  gyration  then 
is  r=  (501  .4/24.06)*  =  4.  56  inches,  and  hence  the  slenderness 
ratio  is  l/r  =  55.3.  Using  for  cj>  the  value  SSWTFJ  tne  formula  now 
gives  5=n  200  pounds  per  square  inch,  so  that  the  factor  of 


ART.  82  SAFE  LOADS  FOR  COLUMNS  205 

safety  is  about  5.3  and  the  column  may  be  regarded  as  having  a 
degree  of  stability  a  little  too  low  for  heavy  traffic. 

The  degree  of  reliability  of  values  of  S  computed  for  columns 
is  very  much  less  than  of  those  computed  for  beams  from  the 
flexure  formula  (41),  since  the  column  formula  has  a  less  reliable 
foundation.  Moreover  it  assumes  that  the  load  is  truly  axial 
and  the  column  perfectly  straight  before  the  application  of  the 
load,  and  these  assumptions  cannot  be  perfectly  realized.  It 
hence  follows  that  factors  of  safety  for  compressive  stress  in 
columns  should  be  higher  than  those  for  beams  and  higher  than 
those  for  direct  compression  on  short  specimens. 

Prob.  81a.  A  cylindrical  wrought-iron  column  with  fixed  ends  is 
12  feet  long,  6.36  inches  in  outside  diameter,  6.02  inches  in  inside 
diameter,  and  carries  a  load  of  49  ooo  pounds.  Find  its  factor  of 
safety. 

Prob.  81b.  A  wooden  stick,  3X4  inches  and  12  feet  long,  is  used 
as  a  column  with  fixed  ends.  Find  its  factor  of  safety  under  a  load  of 
4  ooo  pounds.  If  the  length  of  the  stick  is  only  one  foot,  what  is 
the  factor  of  safety? 

ART.  82.     SAFE  LOADS  FOR  COLUMNS 

To  determine  the  safe  axial  load  for  a  column  of  given  length 
and  cross-section,  it  is  necessary  to  assume  the  allowable  work- 
ing unit-stress  S.  Then  Rankine's  formula  gives, 


in  which  $  is  to  be  taken  from  Art.  80,  and  the  slenderness  ratio 
l/r  is  to  be  computed  from  the  given  data. 

For  example,  let  it  be  required  to  determine  the  safe  load 
for  a  fixed-ended  timber  column,  3X4  inches  in  size  and  10 
feet  long,  so  that  the  greatest  compressive  unit-stress  5  may  be 
800  pounds  per  square  inch.  Hence  a  =12  square  inches,  /  = 
120  inches,  r2  =  4X?,3/i2Xi2  =  f,  and  l2/r2  =  19  200;  also  <£  = 
•g-jnnr-  Then  the  formula  gives  P  =  i  300  pounds  for  the  safe  load. 
A  short  column  of  this  size  should  safely  carry  P=  12X800  = 
9  600  pounds,  or  seven  times  as  much  as  one  of  10  feet  length 


206  COLUMNS  OR  STRUTS  CHAP,  ix 

As  a  second  example  let  it  be  required  to  find  the  axial  load 
for  a  fixed-ended  steel  column  23  feet  6  inches  long  so  that  S  may 
be  12  ooo  pounds  per  square  inch.  Let  the  section  be  that  in 
Fig.  76c,  the  plate  being  }Xi6  inches  and  each  channel  12  inches 
deep  and  weighing  20  J  pounds  per  linear  foot.  From  the  prin- 
ciples of  Arts.  42  and  43,  with  the  help  of  Table  9,  the  least 
radius  of  gyration  of  the  section  is  found  to  be  ^  =  4.56  inches, 
so  that  the  slenderness  ratio  is  //r  =  6i.8.  Then  Rankine's 
column  formula  gives 

P=  24.06X12  ooo  /( iH -1  =  250  500  pounds 

/  \       25  ooo/ 

which  is  the  safe  load  for  the  given  section.  By  using  heavier 
channels  of  the  same  depth  a  much  greater  load  may  be  carried 
without  changing  the  outside  dimensions  of  the  section. 

Prob.  82a.  Find  the  safe  steady  load  for  a  hollow  cast-iron  column 
with  fixed  ends,  the  length  being  18  feet,  outside  dimensions  4X5 
inches,  inside  dimensions  3X4  inches. 

Prob.  S2b.  Find  the  safe  load  for  the  above  steel  column  when  the 
channels  are  12  inches  deep  and  weigh  40  pounds  per  linear  foot,  all 
other  dimensions  and  requirements  remaining  the  same. 

ART.  83.     DESIGN  OF  COLUMNS 

When  a  column  is  to  be  selected  or  designed  the  axial  load  P 
will  be  given,  as  also  its  length  and  the  condition  of  the  ends. 
A  proper  allowable  unit-stress  S  is  assumed,  suitable  for  the 
given  material  under  the  conditions  in  which  it  is  used,  or  the 
value  of  S  will  be  given  in  the  specifications  under  which  the 
design  is  to  be  made.  Then  from  formula  (i),  the  section  area 
of  a  short  column  or  prism  is  P/S,  and  it  is  certain  that  a  greater 
section  area  will  be  needed  for  the  column.  Next,  let  a  cross- 
section  be  assumed,  bearing  in  mind  that  it  will  be  more  effective 
the  further  the  material  be  removed  from  the  axis  (Art.  77). 
For  this  assumed  cross-section  a  and  r  are  to  be  determined,  and 
then  5  is  to  be  computed  from  the  column  formula.  If  this 
computed  value  agrees  with  the  unit-stress  assumed  or  specified, 
a  section  has  been  designed  which  satisfies  the  conditions;  if 


ART.  83  DESIGN   OF   COLUMNS  207 

not,  a  new  cross-section  is  to  be  assumed  and  S  be  again  com- 
puted; this  process  is  to  be  continued  until  a  satisfactory  agree- 
ment is  secured. 

For  example,  a  hollow  cast-iron  rectangular  column,  with 
fixed  ends  and  18  feet  in  length,  is  to  carry  a  load  of  60  ooo  pounds 
and  the  allowable  unit-stress  S  is  to  be  15  ooo  pounds  per  square 
inch.  For  a  short  length  the  area  required  would  be  four  square 
inches;  assume  then  that  about  6  square  inches  will  be  needed. 
Let  the  section  be  square,  the  outside  dimensions  6x6  inches, 
and  the  inside  dimensions  5^X5^  inches.  Then  0  =  5.75  square 
inches,  I=2i6  inches,  ^  =  5.52  inches2,  l/r  =  g2,  and  0  =  7Tnnr- 
Substituting  these  in  Rankine's  formula,  there  is  found  for  ,5 
about  30  ooo  pounds  per  square  inch,  which  is  double  the  speci- 
fied value,  and  hence  the  assumed  dimensions  are  much  too 
small.  Again,  assume  the  outside  dimensions  as  6x6  inches 
and  the  inside  dimensions  as  5  X  5  inches.  Then  a  =  1 1  square 
inches,  7-2  =  5.08  inches2,  and  l/r  =  g6.  Substituting  these  in  the 
formula,  there  is  found  for  S  about  15  700  pounds  per  square 
inch.  Since  this  is  very  near  the  required  working  stress,  it  appears 
that  these  dimensions  very  nearly  satisfy  the  imposed  conditions. 
Many  other  sections  can  also  be  found  which  will  satisfy  the 
requirements,  and  the  one  to  be  finally  selected  will  be  that 
which  is  most  convenient  and  economical. 

In  some  instances  it  is  possible  to  assume  all  the  dimensions 
of  the  column  except  one,  and  then  after  expressing  a  and  r  in 
terms  of  this  unknown  quantity,  to  introduce  them  into  (80)  and 
solve  the  problem  by  finding  the  root  of  the  equation  thus  formed. 
For  example,  let  it  be  required  to  find  the  size  of  a  square  wooden 
column  with  fixed  ends  and  24  feet  long  to  sustain  a  load  of 
100  ooo  pounds  with  a  factor  of  safety  of  10.  Let  x  be  the  unknown 
side ;  then  a  =  x2  and  r2  =  -f?x2,  and  the  column  formula  becomes, 


100  000  / 

800= g —  (T  + 


3 

By  reduction  this  leads  to  the  biquadratic  equation, 

8x4  —  i  ooo#2=33i  776 
and  its  solution  gives  16.6  inches  for  the  side  of  the  column. 


208  COLUMNS  OR  STRUTS  CHAP.  IX 

In  designing  columns  and  beams,  considerations  of  economy 
are  to  be  constantly  kept  in  mind.  For  any  given  data,  it  is 
usually  possible  to  arrange  a  large  number  of  sections  which 
will  satisfy  the  requirements  regarding  strength,  and  the  most 
advantageous  one  of  these  is  that  which  can  be  built  at  the  low- 
est possible  cost.  In  architecture,  considerations  of  beauty 
are  also  to  be  followed  in  order  that  the  eye  may  be  pleased  with 
the  view  of  the  column.  It  is  sometimes  said  that  beautiful 
forms  are  those  of  greatest  strength,  and  this  now  and  then  hap- 
pens to  be  the  case,  but  beauty  and  economy  are  often  con- 
tradictory elements. 

Prob.  83.  Compute  the  size  of  a  square  wooden  column,  12  feet 
long  and  having  fixed  ends,  to  carry  an  axial  load  of  50  net  tons  with 
a  factor  of  safety  of  TO.  Compute  also  the  size  of  the  column  for  the 
case  of  round  ends. 

ART.  84.    THE  STRAIGHT-LINE  FORMULA 

The  column  formula  (80)  was  derived  by  Rankine  about 
1860  from  older  forms  deduced  by  Tredgold  and  by  Gordon, 
in  which  the  ratio  l/d  was  used,  d  being  the  least  thickness  of 
a  rectangular  section  or  the  diameter  of  a  circular  section ;  Ran- 
kine introduced  the  ratio  l/r  and  thus  produced  a  formula  appli- 
cable to  all  kinds  of  cross-sections.  This  formula  has  been  more 
widely  used  than  any  other  notwithstanding  its  empirical  nature. 
When  it  is  compared,  however,  with  the  results  of  many  experi- 
ments, it  is  seen  that  in  many  cases  these  results  may  be  repre- 
sented by  a  straight  line,  within  the  limits  of  the  ratio  l/r  gen- 
erally used,  almost  as  well  as  by  the  curve  of  Rankine's  formula. 
On  this  basis  the  straight-line  formula  for  columns  was  first 
deduced  in  1886  by  T.  H.  Johnson. 

On  Fig.  84  are  shown  fifteen  points  which  represent  the  aver- 
age results  of  about  sixty  experiments  made  by  Tetmajer  on 
struts  of  medium  steel  of  different  lengths  and  sizes,  the  ordinate 
for  each  point  giving  the  unit-load  P/a  which  caused  the  failure 
of  the  struts  having  the  slendemess  ratio  l/r  corresponding  to 
its  abscissa.  The  broken  curve  is  that  of  Euler's  formula  and 


ART.  84 


THE  STRAIGHT-LINE  FORMULA 


209 


it  is  seen  to  fit  the  observations  very  well  for  values  of  l/r  greater 
than  150.  For  lower  values  of  l/r  the  full  straight  line  seems 
to  give  a  fair  average  representation  of  the  observations,  this 
being  drawn  tangent  to  Euler's  curve.  It  is  required  to  deter- 
mine the  equation  of  this  straight  line. 

Let  y  be  the  ordinate  P/a  and  x  the  abscissa  l/r  and  5  be 
the  value  of  P/a  when  x  is  zero,  or  the  distance  from  the  origin 
to  the  point  where  the  straight  line  cuts  the  axis  of  ordinates. 
Then  the  equations  of  Euler's  curve  and  of  the  straight  line  are, 

y=fiE/x?  and  y=S-Cx 

in  which  the  parameter  C  is  to  be  determined  by  making  the 
straight  line  tangent  to  the  curve.  By  equating  the  values  of  y 
in  these  two  equations  and  also  the  values  of  the  first  derivatives 
dy/dx,  the  ordinate  and  abscissa  of  the  point  of  tangency  are 
found  to  be, 

i  =  $S  and  xi 


Inserting  these  in  the  equation  of  the  straight  line,  the  value  of 
C  is  found,  and  accordingly  may  be  written, 


and 


(84) 


a  r 

in  which  the  number  fj.  is  rc2  for  round  ends,  2.o$7t2  for  one  end 
round  and  the  other  fixed,  and  4?r2  for  fixed  ends  (Art.  78). 


40006 
^30-000 
°20000 

V 

a 

"X 

\ 

X 

> 

\ 

\ 

v  — 

0 
Ss 

0 

—  „ 

\ 

-^\ 

x 

^o 

W* 

?s 

Q. 

f^l&OOO 
jO 

•—  • 

"""" 

0- 

-0 

- 

.--. 

60  100  150 

Values  of  the  ratio  l/r 

FlG.  84 


200 


The  values  of  S  to  be  used  for  cases  of  rupture  are  such  as 
to  make  the  straight  line  agree  best  with  experimental  results. 
The  values  derived  by  Johnson  are  given  in  the  following  table, 
together  with  his  values  of  C  and  the  limiting  values  of  l/r  corre- 


210 


COLUMNS  OR  STRUTS 


CHAP.  IX 


spending  to  the  point  of  tangency.  The  above  theoretic  values 
of  /JL  were  not  used  in  computing  C,  as  his  experiments  indicated 
that  the  numbers  n2,  i§rc2,  and  2j;r2  for  round,  hinged,  and  flat 
ends  respectively  gave  a  closer  agreement.  It  will  be  noticed 
CONSTANTS  FOR  FORMULA  (84) 


Kind  of  Column 

Pounds  per  Square  Inch 

Limit  of 
l/r 

5 

C 

Wrought  iron 

Flat  ends 

42  ooo 

128 

218 

Hinged  ends 

42  ooo 

157 

178 

Round  ends 

42  ooo 

203 

138 

Structural  steel 

Flat  ends 

52  5°° 

179 

I9S 

Hinged  ends 

52  5oo 

220 

159 

Round  ends 

52500 

284 

123 

Cast  iron 

Flat  ends 

80  ooo 

438 

122 

Hinged  ends 

80  ooo 

537 

99 

Round  ends 

80  ooo 

693 

77 

Oak 

Flat  ends 

5  400 

28 

128 

that  the  values  of  S  in  the  table  are  less  than  the  average  ulti- 
mate compressive  strengths  given  in  Art.  5.  For  ductile  mate- 
rials like  wrought  iron  and  structural  steel,  this  should  be  the 
case  in  columns,  since  when  the  elastic  limit  is  passed  a  flow 
of  metal  begins  which  causes  the  lateral  deflection  to  increase, 
and  failure  then  rapidly  follows. 

The  straight-line  formula  is  not  suitable  for  investigating 
a  column,  that  is,  for  determining  values  of  S  due  to  given  loads, 
because  S  enters  the  formula  in  such  a  manner  as  to  lead  to  a 
cubic  equation  when  it  is  the  only  unknown  quantity.  It  may 
be  used  to  find  the  safe  load  for  a  given  column  to  withstand 
a  given  unit-stress  S,  or  to  design  a  column  for  a  given  load  and 
unit-stress.  When  so  used,  it  is  customary  to  divide  the  values 
of  S  and  C  given  in  the  table  by  an  assumed  factor  of  safety. 
For  example,  Cooper's  specifications  require  that  the  section 
area  a  for  a  medium -steel  post  of  a  through  railroad  bridge  shall 


A.RT.  85  OTHER    COLUMN   FORMULAS  211 

be  found  from  P/a  =  17  000  —  90  (//>)  pounds  per  square  inch, 
in  which  P  is  the  direct  dead-load  compression  on  the  post  plus 
twice  the  direct  live-load  compression;  the  values  of  S  and  C 
here  used  are  a  little  less  than  one-third  of  those  given  in  the 
table  for  round  ends. 

While  the  straight-line  formula  is  sometimes  slightly  more 
convenient  than  that  of  Rankine,  it  cannot  be  regarded  either 
as  having  so  high  a  degree  of  validity  or  as  satisfying  so  well 
the  results  of  experiments.  It  is  hence  advisable  that  the  use 
of  this  formula  should  be  limited  to  cases  in  which  specifica- 
tions require  it  to  be  employed,  and  for  rough  approximate 
computations. 

Prob.  84.  Solve  Probs.  82a  and  83  by  the  help  of  the  straight-line 
formula,  applying  the  factors  of  safety  to  the  values  of  S  and  C  given  in 
the  above  table. 

ART.  85.     OTHER  COLUMN  FORMULAS 

Many  attempts  have  been  made  to  establish  a  formula  for  col- 
umns which  shall  be  theoretically  correct,  like  the  flexure  formula 
(41)  for  beams,  when  the  material  is  not  stressed  beyond  the 
elastic  limit.  Although  many  column  formulas  have  been  proposed 
vvhich  have  been  claimed  by  their  authors  to  have  a  rational 
basis,  none  of  them  has  yet  been  recognized  by  the  engineering 
profession  as  more  satisfactory  than  the  formula  of  Rankine. 
For  long  columns  all  agree  that  Euler's  formula  is  correct  in 
giving  the  load  which  causes  failure  by  lateral  bending,  but  for 
the  columns  commonly  used  in  practice,  where  the  slenderness 
ratio  l/r  lies  between  30  and  150,  a  fully  satisfactory  formula 
has  not  been  established. 

In  1873  Ritter  proposed  a  formula  to  be  used  when  the  unit- 
stress  S  does  not  exceed  the  elastic  limit  Se,  namely, 
P  S 


in  which  P/a  is  the  axial  unit-load,  E  the  modulus  of  elasticity 
of  the  material,  and  the  number  fi  is  n2  for  round  ends,  2.o$x2 
for  one  end  round  and  the  other  fixed,  and  ^  for  both  ends 


Ends 
Fixed 

Fixed  and 
Round 

Ends 
Round 

I 

i-9S 

4 

20  ooo 

20000 

20  ooo 

I 

T-95 

4 
30000 

30000 

30000 

I 

i-95 

4 

40  ooo 

40000 

40  ooo 

I 

34000 

i-95 
34000 

4 

34000 

212  COLUMNS  OR  STRUTS  CHAP.  IX 

fixed.  This  is  seen  to  be  the  same  in  form  as  Rankine's  formula, 
the  constant  <£  having  the  value  Sf/fiE.  Using  the  average 
values  of  S,  and  E  given  in  Arts.  2  and  9,  the  values  of  cf>  for 
columns  of  different  materials  are, 


for  Timber 
for  Cast  Iron 
for  Wrought  Iron 
for  Structural  Steel 


These  values  of  <£  are  smaller  than  the  empirical  ones  in  Art.  80, 
and  hence  the  formula  of  Ritter  gives  larger  values  of  P/a  than 
the  formula  of  Rankine.  When  l/r  is  100,  the  values  of  P/a  for 
fixed-ended  columns  found  from  Ritter's  formula  are  2  percent 
greater  than  those  given  by  Rankine's  formula  for  wrought  iron, 
8  percent  greater  for  structural  steel,  230  percent  greater  for 
cast  iron,  and  290  percent  greater  for  timber.  While  the  com- 
parison is  fairly  satisfactory  for  wrought  iron  and  medium  steel, 
the  disagreement  in  the  results  for  cast  iron  and  timber  is  so 
great  that  Ritter's  formula  cannot  be  regarded  as  satisfactory. 

Another  formula  which  has  received  extended  discussion  is 
that  derived  by  Crehore  in  1879  and  also  independently  deduced 
by  Reuleaux,  Marburg,  and  others.  This  formula  is  identical 
with  that  of  Ritter  except  that  Se  is  replaced  by  S\  designating 
P/a  by  B,  it  may  be  written, 

*?  T> 

B~i+.(S/rLE)(l/r?  S    i-(B/pE)(l/')2 

from  the  first  of  which  the  unit-load  may  be  computed  for  a 
given  unit-stress,  while  the  second  gives  the  compressive  unit- 
stress  on  the  concave  side  of  the  column  due  to  a  given  unit-load. 
Tables  giving  values  of  S  for  wrought  iron  and  steel  columns 
were  published  by  Merriman  in  1894,  together  with  a  deriva- 


ART.  85  OTHER   COLUMN   FORMULAS  213 

tion  different  from  that  of  Crehore.  These  tables  show  that 
values  of  P/a  computed  for  a  given  unit-stress  S  are  greater 
than  those  found  from  Rankine's  formula;  for  instance,  when 
l/r=ioo  and  S  =12  ooo  pounds  per  square  inch,  it  gives 
P/a=n  coo  for  steel  columns  with  fixed  ends,  whereas  Rankine's 
formula  gives  P/a  =  8  500  pounds  per  square  inch.  In  general 
the  formula  gives  working  unit-loads  which  are  from  20  to  30 
percent  greater  than  those  derived  from  the  expression  of  Rankine 
and  hence  it  cannot  at  present  be  regarded  with  favor.  On 
the  other  hand  when  applied  to  cases  of  rupture,  it  usually  gives 
values  of  P/a  less  than  those  shown  from  experiment  and  for 
this  reason  it  is  also  unsatisfactory. 

The  discussions  of  this  chapter  show  that  the  theory  of  columns 
stands  upon  a  less  rational  basis  than  that  of  beams.  The  flexure 
formula  S  .  I/c  =  M  has  a  sound  theoretical  foundation  and, 
for  all  cases  where  the  elastic  limit  is  not  exceeded,  it  is  found 
to  agree  with  experiment.  Euler's  formula  for  long  columns 
has  a  sound  theoretical  basis  and  it  also  agrees  with  experiment, 
but  all  other  column  formulas  contain  certain  theoretic  defects. 
A  rational  formula  can  often  be  safely  applied  to  cases  of  rupture 
by  using  empirical  constants,  but  it  is  rarely  the  case  that  an 
empirical  formula  containing  constants  deduced  from  experi- 
ments on  rupture  can  be  applied  to  cases  in  which  the  elastic 
limit  is  not  exceeded,  except  by  the  use  of  arbitrary  factors  of 
safety.  For  this  reason  Rankine's  formula  is  not  a  satisfactory 
one,  and  yet  the  long  use  of  it  by  the  engineering  profession  has 
built  up  a  system  of  practice  and  precedent  which  must  con- 
tinue to  be  respected  until  the  time  arrives  when  a  satisfactory 
theoretical  formula  shall  be  established.  The  present  indica- 
tions are  that  the  errors  of  Rankine's  formula,  when  used  with 
the  average  empirical  values  of  <p  given  in  Art.  80,  are  on  the 
side  of  safety. 

Prob.  85a.  Refer  to  Van  Nostrand's  Engineering  Magazine  for 
December,  1879,  and  ascertain  the  assumption  used  by  Crehore  in  his 
derivation  of  the  above  column  formula.  Compare  also  the  discussion 
of  the  modified  Euler  formula  in  Art.  167. 


214 


COLUMNS  OR  STRUTS 


CHAP.  IX 


ART.  86.    ECCENTRIC  LOADS  ON  PRISMS 

In  all  the  preceding  discussions,  the  load  on  the  column 
has  been  regarded  as  axial,  but  cases  are  very  common  in  engineer- 
ing practice  where  the  load  P  is  applied  at  a  distance  p  from 
the  axis  of  the  column. ;  this  lever  arm  p  is  called  the  "eccentricity  " 
of  the  load.  It  is  evident  that  the  flexural  stresses  in  the  column 
will  increase  with  p  and  that  the  total  unit-stress  S  on  the  most 
compressed  side  of  the  column  will  be  greater  than  for  the  case 
of  an  axial  load. 


Fig.  86a 

Fig.  86a  shows  two  rectangular  columns  with  eccentric  loads. 
In  the  first  diagram,  the  weight  P  on  the  top  has  a  resultant 
shown  by  the  arrow  which  falls  within  the  cross-section;  in 
the  second,  this  resultant  line  falls  without  the  cross-section.  In 
these  two  diagrams,  the  point  of  application  of  P  is  on  a  median 
line  of  the  cross-section,  and  this  is  the  common  case  in  prac- 
tice, but  Fig.  86b  shows  the  unusual  case  where  the  point  of  appli- 
cation is  not  on  a  median  line' of  the  cross-section  of  the  column. 
In  all  cases  like  those  of  Fig.  86a,  however,  the  tendency  of 
the  load  is  to  cause  rotation  about  an  axis  perpendicular  to  direc- 
tion of  the  lever  arm  p.  Thus,  in  each  of  the  figures,  let  mn  be 
drawn  in  the  plane  of  the  cross -section  through  the  axis  of  the 
column  and  normal  to  p;  then  the  flexural  stresses  are  to  be 
referred  to  mn  as  a  neutral  axis,  and  the  moment  of  inertia 


ART.  86  ECCENTRIC    LOADS   ON   PRISMS  215 

of  the  cross-section  with  respect  to  this  axis  mn  is  to  be  deter- 
mined by  the  methods  of  Art.  43.  The  shaded  areas  in  Fig.  86a 
show  the  compressive  unit-stresses  due  to  the  eccentric  load 
when  the  elastic  limit  of  the  material  is  not  exceeded,  the  varia- 
tion occurring  at  a  uniform  rate. 

Considering  first  the  cases  of  Fig.  86a,  where  the  load  P  is 
on  a  median  line  of  the  section  area  a,  the  total  compressive 
stress  on  this  area  must  be  also  P,  otherwise  equilibrium  could 
not  obtain,  and  hence  the  average  compressive  unit-stress  is 
P/a,  which  is  represented  by  cd  in  the  diagrams.  On  the  most 
compressed  side  this  is  increased  to  as  and  on  the  other  side 
it  is  decreased  to  U,  the  triangles  pds  and  qdt  representing  the 
flexural  unit-stresses  exactly  as  in  the  discussion  of  Art.  80.  Let 
the  maximum  unit-stress  as  be  denoted  by  S  and  the  part  ps 
due  to  the  flexure  be  denoted  by  S^  then  S  =  P/a-\-Sl.  From 
the  flexure  formula  (41)  Sl  =  Mc/I  where  M  is  the  bending 
moment  Pp^  c  is  the  distance  ca  from  the  axis  of  the  prism  to  the 
most  compressed  side,  and  /  is  the  moment  of  inertia  of  the 
section  area  a  with  respect  to  the  axis  mn.  Now  I  =  ar2  where 
r  is  the  radius  of  gyration  of  the  section  area  with  respect  to  mn. 
Hence  Sl=  Ppc/ar2  is  the  flexural  unit  stress  represented  by  ps 
in  the  diagram.  Inserting  this  in  the  above  expression  for  S 
there  is  found 

or    ^=-7-,  (86) 

a  2 


from  which  S  can  be  computed  for  a  given  P/a,  or  P/a  be  found 
for  a  given  5.  This  formula  is  only  valid  when  the  unit-stress 
S  does  not  exceed  the  elastic  limit  of  the  material. 

When  the  section  area  is  a  rectangle  having  the  depth  d 
in  a  direction  through  P  and  the  axis  of  the  prism,  the  value  of 
r2  is  ^bd2/bd  =  -^d2,  and  since  c  is  $d  for  this  case,  the  expres- 
sion for  S  becomes 


which  is  the  same  result  as  found  in  Art.  29.     For  a  circle  of 
diameter  d  the  value  of  r2  is  r^2  and  that  of  c  is  \d,  hence 

S=(P/a)(i+Sp/d). 


216  COLUMNS  OR  STRUTS         CHAP.  IX 

Let  c'  be  the  distance  from  the  axis  of  the  prism  to  the  side 
where  the  unit-stress  is  bt  in  the  diagram.  Then  this  unit- 
stress  is  S'=(P/a)(i  -c'p/r2),  and  Sf  will  be  tension  when  p 
is  greater  than  r2/c'.  This  case  is  shown  in  the  second  diagram 
of  Fig.  86#,  where  the  point  e  is  within  the  section  area. 

Second,  let  the  case  of  Fig.  866  be  considered,  where  the 
load  P  is  not  applied  on  a  median  line  of  the  section  area.  Let 
i-i  and  2-2  be  two  median  lines  of  the  section,  or  the  two  prin- 
cipal axes  about  which  the  greatest  and  least  moments  of  inertia 
of  the  section  occur.  Let  p^  and  p2  be  the  eccentricities  of  P 
with  respect  to  these  axes,  and  let  x  and  y  be  the  coordinates 
of  any  point  of  the  section  area  with  respect  to  the  same  axes. 
Here  the  load  P  acting  with  the  eccentricity  pv  produces  a  unit- 
stress  6\  at  the  point  whose  coordinates  are  x  and  y,  while  P 
acting  with  the  eccentricity  p2  produces  a  unit-stress  S2  at  the 
same  point.  The  average  unit-stress,  as  before,  is  P/a;  hence 
the  total  unit-stress  at  the  given  point  is  S  =  P/a+S1  +  S2.  Now 
S^MyCjI  and  S2  =  M2c2/Ii)  where  Ml  =  Pp1  and  M2  =  Pp2, 
cl  =  oc,  and  c2  =  y,  and  Il  =  ar12)  and  I2  =  ar22  are  the  moments  of 
iaertia  of  the  section  area  with  respect  to  the  axes  i-i  and  2-2, 
where  rl  and  r2  are  the  principal  radii  of  gyration.  Hence, 


00 

a  \       ri2     r2 

is  the  compressive  unit-stress  at  the  given  point  due  to  the  eccen- 
tric load.     This  formula  applies  to  a  section  area  of  any  shape. 

For  the  case  of  the  rectangle  A  BCD  shown  in  Fig.  866,  let 
AB  =  d  and  AD  =  b,  and  let  it  be  required  to  find  the  compressive 
unit-stresses  at  the  four  corners  due  to  P.  Here  rf  =  d2/i2  and 
r22  =  b2/i2.  For  the  corner  A  the  value  of  x  is  +d/2  and  that 
of  y  is  +6/2,  and  accordingly 

SA  =  (P/a)(*+6pi/d+6p2/b). 
For  the  corner  B,  x  is  -d/2  and  y  is  +6/2,  hence 

SB=(P/a)(i-6pi/d+6p2/b). 
For  the  corners  C  and  D  are  also  found 

Sc=(P/a)(i-6pi/d-6p2/b),    SD=(P/a)(i  +  6p1/d-6p2/b). 
If,  in  computation,  one  or  more  of  these  stresses  is  found  to  be 
negative,  it  indicates  tension  instead  of  compression. 


ART.  87 


ECCENTRIC  LOADS  ON  COLUMNS 


217 


Prob.  86a.  In  Fig.  83a  let  the  di  tance  ce  be  denoted  by  z.     Prove 
that  the  product  pz  equals  r 2  for  both  diagrams. 

ART.  87.     ECCENTRIC  LOADS  ON  COLUMNS 

The  investigation  of  the  last  article  gives  results  for  5  which 
are  too  small,  especially  for  columns  having  a  slenderness  ratio 
greater  than  about  100,  since  the  lateral  deflection  of  the  column 
due  to  the  eccentric  load  will  increase  the  eccentricity  of  the 
load  for  all  sections  except  those  at  the  ends.  Thus,  in  the  fol- 
lowing figures,  let  p  be  the  eccentricity  of  the  load  P,  and  /  the 
maximum  deflection  of  the  axis  of  the  column  from  its  original 
straight  position ;  then  the  lever  arm  or  eccentricity  of  the  load 
with  respect  to  the  neutral  axis  of  the  dangerous  section  is 
which  may  be  called  q,  so  that  formula  (86)  becomes, 


and  hence  /  or  q  must  be  determined  in  order  to  find  S. 

f\~*-i  PU  PH 


*\ 

Fig.  87a  Fig.  876  Fig.  87c 

Let  PO  be  the  axial  load  given  by  Euler's  formula  for  long 
columns,  and  suppose  that  this  load  is  applied  to  the  column 
when  it  has  the  deflection  /,  the  eccentric  load  P  being  removed. 
Then  this  load  P0  will  hold  the  long  column  in  equilibrium  and 
its  moment  with  respect  to  the  axis  of  the  dangerous  section  is 
PO/,  while  the  moment  of  the  eccentric  load  with  respect  to  the 
same  section  was  P(p+f).  Equating  these  moments,  it  follows 
that, 

f=Pp/(P0~P)  and 


which  indicates  that  the  deflection  of  a  column  under  an  eccentric 
load  is  determinate,  although  under  an  axial  load  it  is  indeter- 


218  COLUMNS  OR  STRUTS  CHAP,  ix 

minate.  Deflections  computed  from  this  formula  are,  however, 
too  small,  because  it  has  been  deduced  by  considering  only  the 
moment  at  the  middle  of  the  column,  whereas  all  moments  should 
be  taken  into  account. 

Let  Fig.  S7b  represent  a  column  of  length  /  which  is  free  at 
the  upper  end  and  vertically  fixed  at  its  lower  end.  Let  the 
origin  of  coordinates  be  taken  at  the  top  under  the  load,  values 
of  x  being  measured  downward  and  those  of  y  horizontally.  The 
bending  moment  for  any  point  of  the  elastic  curve  is  —Py,  and 
hence  El .  d2y/dx2  =  —  Py,  in  which  I  is  moment  of  inertia  of 
the  cross-section  and  E  is  the  modulus  of  elasticity  of  the  mate- 
rial. Integrating  this  and  determining  the  constant  by  the  con- 
dition that  the  tangent  dy/dx  =  o  when  y  =  q,  there  results 
El. (dy / doc)2  =P(q2—y2).  Integrating  again  and  ascertaining  the 
constant  by  the  condition  that  y  =  p  when  x  =  o,  there  is  found, 

arc  sir/=  (P/E/)*#+arc  sin^ 

for  the  equation  of  the  elastic  curve.  From  this  equation  the 
total  eccentricity  q  is  determined  by  making  x  =  l  for  y  =  gyand, 

p=q  sin[^-(P/£/)*7]         whence  q=p  sec(P/EI)*l 

Now  let  6  represent  the  number  (P12/EI)*,  this  being  an  angle 
measured  in  radians.  Then  the  value  of  q  is  given  by, 

q=p  sec0=/<i  +  o.502  +  o.2o86>4+o.o84706+.  .  .)  (87)' 

where  the  quantity  in  the  parenthesis  is  obtained  by  expanding 
sed?  into  a  series  by  Maclaurin's  theorem. 

By  referring  to  Art.  78  it  will  be  seen  that  the  above  value 
of  0  applies  to  a  column  with  two  round  ends  if  /  is  replaced  by 
J/,  to  a  column  with  one  end  round  and  the  other  fixed  if  /  is  re- 
placed by  0.35/5  and  to  a  column  with  two  fixed  ends  if  /  is  replaced 
by  i/.  Accordingly  in  (87)  let  the  unit-load  P/a  be  represented 
by  B]  also  in  the  value  of  6  let  /  be  replaced  by  ar2,  where  r  is 
the  least  radius  of  gyration  of  the  section.  Then  the  formula, 

S=B(I+^  secfl)  ==J3+J^|-(i+o.502+o.2o804+.  .  .)       (87)" 
applies  to  all  columns,  when  62  has  the  following  values: 


ART.  87  ECCENTRIC    LOADS    ON   COLUMNS  219 

for  one  end  free  and  the  other  fixed,      62  =  B/E  .  (l/rf 
for  both  ends  round,  62=\B/E  .  (l/r)2 

for  one  end  round  and  the  other  fixed,  62=~8B/E  .  (l/r)2 
for  both  ends  fixed,  62  =  &B/E  .  (l/r)2 


In  computing  the  unit-stress  5,  the  value  of  sec  0  may  be 
found  for  all  cases  by  the  help  of  Table  17,  while  the  series 
cannot  be  used  unless  6  is  less  than  \K  :  when  O2  equals  fa2 
the  unit  load  B  is  that  given  by  Euler's  formula  and  S  is  in- 
finite. A  closely  approximate  value  of  sec  Q  is  given  by 


For  example,  let  the  unit-load  B  be  10  ooo  pounds  per  square 
inch,  applied  at  a  distance  of  i.oi  inches  from  the  axis  of  a  steel 
column  having  round  ends,  the  length  of  the  column  being  192 
inches,  the  distance  c  being  4.45  inches,  and  the  radius  of  gyra- 
tion in  the  direction  of  the  eccentricity  being  3.00  inches.  Hence 
02  =  \(B/E)(l/r)2  =  0.3413,  and  then  the  series  in  (87)"  gives 
5=16900  pounds  per  square  inch,  so  that  the  eccentric  load 
increases  the  mean  unit-stress  about  69  percent.  Or,  using  the 
value  of  6,  formula  (87)'  gives  (7=1.197  inches  and  then  from 
(87)  the  unit-stress  5  is  directly  found. 

To  illustrate  the  computation  of  seed  and  at  the  same  time 
show  the  great  influence  of  an  increase  in  length  of  the  column, 
take  the  same  data  as  above  except  that  the  length  is  twice  as 
great,  or  384  inches.  Then,  #2  =  1.3653  and  #  =  1.168  radians  = 
65°  45',  whence  from  a  trigonometric  table  sec/9  =  1.934.  Accord- 
ingly the  total  eccentricity  is  5  =  1.934^  =  1.953  inches,  and  the 
unit-stress  S  is  20  700  pounds  per  square  inch. 

When  a  column  is  to  be  designed,  its  length  and  eccentric 
load  are  given,  as  also  the  allowable  unit-stress  5,  and  such  values 
of  a,  c,  and  r  are  to  be  found  as  will  satisfy  (87)"  and  at  the  same 
time  give  the  greatest  economy.  This  process  is  a  tentative  one, 
and  several  trials  may  be  necessary  in  order  to  find  a  satisfactory 
section.  Probably  the  best  plan  is  to  assume  values  of  a,  c,  and  r 
and  then  compute  5,  continuing  the  process  until  the  computed 
allowable  values  fairly  agree. 


220  COLUMNS  OR  STRUTS  CHAP.  IX 

The  above  formula  (87)"  is  not  convenient  for  the  direct 
computation  of  the  eccentric  unit-load  P/a  for  a  given  column 
under  a  given  unit-stress  5,  and  hence  a  problem  of  this  kind 
is  also  to  be  solved  by  successive  trials,  values  of  P/a  or  B  being 
inserted  in  the  formula  until  one  is  found  which  makes  S  the 
same  as  the  given  value.  This  process,  as  also  that  of  designing 
a  column  section,  will  be  often  facilitated  by  assuming  sec#  for 
the  first  trial,  taking  it  as  unity  if  the  column  is  short,  or  about 
i  £  or  2  for  a  long  column. 

Prob.  87 a.  Show  that  (p+f)c  equals  r2  for  a  column  so  deflected 
that  there  is  no  stress  on  the  convex  side. 

Prob.  87b.  A  wooden  strut  with  fixed  ends  is  18  feet  long,  4  inches 
square,  and  the  compression  of  5  ooo  pounds  is  applied  half-way  be- 
tween the  center  and  corner  of  the  end  sections.  Compute  the  deflec- 
tion at  the  middle  of  the  column,  and  the  maximum  unit-stress  S. 

Prob.  S7c.  Prove  that  the  eccentric  load  P  which  holds  a  round- 
ended  column  in  equilibrium  with  the  deflection  q—p,  is  given  by  the 
formula  P/a=4E(r/l)2(snc  cosp/q)2. 


ART.  88.     ON  THE  THEORY  OF  COLUMNS 

Since  the  discussion  in  Art.  78  does  not  determine  Euler's 
formula  for  a  column  with  one  end  round  and  the  other  fixed,  a 
more  general  investigation  will  now  be  given.  Let  the  column 
be  constrained  at  the  ends  so  that  bending  moments  M\  and 
MI  exist  there  when  a  lateral  deflection  occurs.  The  bending 
moment  at  any  section  distant  x  from  the  end  i  of  the  column  is 
M  =  M\  +  Vx  —  Py  where  Vi  is  the  transverse  shear  at  i.  When 
x  =  l,  then  y=--o  and  M  =  M2,  so  that  Vi  =  (M2-Mi)/L  The 
general  equation  of  the  elastic  curve  now  is 


and  the  general  solution  of  this  differential  equation  is 

p— 


(88) 


ART.  88  ON    THE   THEORY  OF    COLUMNS   .  221 

in  which  A,  B,  C,  D,  /?  are  constants  of  integration  to  be  found 
from  the  condition  of  the  problem.  Differentiating  (88)  twice, 
there  result, 


g=  £  (A  cos/?y 


+C~D 


Comparing  the  two  values  for  EI.82y/dx2  and  eliminating  the 
trigonometric  expression  by  the  help  of  (88),  three  constants  are 
found,  namely,  p2  =  Pl2/EI,  C  =  M2l2/p2,  and  D  =  Mll2/p2. 
Further,  the  first  member  of  (88)  "  becomes  MI  when  #  =  0  and 
M2  when  x  =  l\  hence  A  =  -M2l2/p2  sin/9  and  B=-Mll2/^2  sind. 
The  expression  found  for  /?2  gives  P  =  82EI/l2  from  which  Euler's 
formulas  may  be  deduced  for  special  arrangements  of  the  ends. 

Inserting  the  values  of  the  constants  in  (88)  '  and  making 
x  =  o  there  is  found  an  expression  for  the  tangent  of  the  angle 
which  the  tangent  to  the  elastic  curve  at  the  end  i  of  the  deflected 
column  makes  with  the  axis  of  x,  namely, 


Now  let  the  end  2  be  round  and  the  end  i  be  fixed,  then  M2  =  o, 
and  since  dy/dx  must  be  zero  for  a  fixed  end,  /?  =  tan/?  is  the 
condition  that  end  i  is  fixed  and  end  2  is  round.  This  equation 
has  two  roots,  the  first  being  /?  =  o  which  applies  to  a  straight 
undeflected  column.  The  other  root  /?  =  4.49341  corresponds 
to  the  least  value  of  P  which  holds  the  deflected  column  in  equi- 
librium. Since  4.49341  =  1.43029^,  the  value  of  /?2  is  2.0457^, 
and  hence  Euler's  formula  for  this  case  is  P  =  2.o46n2EI  /I2,  as 
noted  in  Art.  78.  The  constant  2j,  which  is  usually  stated  for 
this  case,  gives  the  value  of  P  about  10  percent  too  large. 

From  (88)"'  the  values  of  /?  for  both  ends  round  or  for  both 
ends  fixed  are  also  readily  obtained.  For  both  ends  round, 
MI  and  M2  are  zero,  while  dy/dx  is  indeterminate.  Hence  the 
quantity  0(1  -cos/?)  /sin/?  must  be  indeterminate,  and  this  will 


222  COLUMNS  OR  STRUTS  CHAP.  IX 

be  secured  by  making  sin/9  =  o,  whence  /?  =  TT  for  the  least  value 
of  P  and  accordingly  Euler's  formula  for  columns  with  round 
ends  is  P  =  x2EI/l2.  When  both  ends  are  fixed  and  M2  equals 
Afi,  then  dy/dx  becomes  zero  for  both  x  =  o  and  x  =  l  provided 
cos5  =  i;  this  requires  that  /?=27r,  whence  Euler's  formula  for 
columns  with  two  fixed  ends  is  P  =  47i2EI /I2. 

Taking  the  moments  at  fixed  ends  as  positive,  the  maximum 
negative  moment  may  be  found  from  (88)"'.  When  both  ends 
are  fixed  the  maximum  negative  moment  is  at  the  middle  of  the 
length  and  is  double  the  positive  moment  at  the  end.  When 
end  2  is  round  and  end  i  is  fixed,  all  moments  have  the  same 
sign  as  MI  and  the  maximum  is  about  i.o2M\. 

An  ideal  column  is  one  which  is  perfectly  straight  in  its 
unstressed  state  and  which  when  laterally  deflected  by  an  extrane- 
ous force  returns  to  its  straight  condition  on  the  removal  of 
that  force.  The  maximum  unit-stress  for  the  ideal  column  hence 
does  not  exceed  the  elastic  limit  Se.  For  the  straight  column 
failure  begins  when  P/a  equals  Se',  for  the  deflected  column 
failure  begins  when  Euler's  condition  P/a  =  /iE(r/l)2  is  reached. 
Hence  Se  =  nE(r/l)2  or  l/r=(uE/Se)*  gives  the  critical  value  of 
l/r  between  the  two  methods  of  failure.  For  instance,  columns 
of  structural  steel  with  round  ends,  for  which  fi.  =  n2,  have  l/r  =  92, 
while  columns  of  structural  steel  with  fixed  ends,  for  which 
,«  =  7r2  have  //r  =  i84  as  critical  values  of  the  slenderness  ratio. 
For  ideal  conditions  Euler's  formula  only  applies  to  higher 
values  of  l/r  than  these  critical  ones.  Under  actual  conditions, 
however,  columns  are  not  perfectly  straight,  so  that  flexure  may 
begin  long  before  P  reaches  the  value  given  by  Euler's  formula. 
In  other  words,  there  is  always  an  initial  deflection  in  practise, 
and  Rankine's  formula  is  an  attempt  to  empirically  take  its  average 
value  into  account. 

The  theoretic  basis  of  Rankine's  formula  seems  far  more 
satisfactory  than  that  of  any  other  which  has  been  proposed  for 
the  discussion  of  such  columns  as  are  used  in  engineering  practice. 
Nevertheless  it  should  be  noted  that  the  constants  given  on 


ART.  88  ON   THE    THEORY    OF   COLUMNS  223 

page  202  cannot  be  expected  to  apply  to  unusual  or  extreme 
cases,  for  they  are  averages  deduced  from  tests  on  common  sec- 
tions. In  this  connection  the  investigations  made  in  Ireland  by 
Lilly  in  1907  may  be  noted  as  important.  Making  tests  upon 
hollow  cylindrical  columns  of  small  thickness,  he  found  failure 
to  occur  under  a  much  less  unit  load  P/a  than  for  solid  cylindrical 
columns  having  the  same  values  of  l/r,  and  it  was  seen  that 
this  was  due  to  a  wave-like  crippling  that  produced  secondary 
flexure.  His  investigations  lead  to  the  conclusion  that  the 
formula  for  such  columns  is 

P=  S 

a 


in  which  t  is  the  thickness  of  the  material,  and  m  and  n  are 
constants  that  depend  upon  the  kind  of  material  and  the  condition 
of  the  ends.  This  expression  indicates  that  the  strength  of  a 
hollow  cylinder  depends  not  merely  upon  its  section  area  and 
slenderness  ratio  but  also  upon  the  thickness  of  its  walls.  When 
t  is  very  small,  Lilly's  formula  makes  P/a  also  small,  and  it  is 
evident  that  this  conclusion  is  a  correct  one. 

When  a  compound  column  is  made  up  of  plates  or  webs, 
as  was  the  case  in  the  lower  chord  of  the  ill-fated  Quebec  bridge, 
which  failed  in  1907,  it  is  also  probable  that  neither  Rankine's 
formula  or  even  the  simple  formula  P/a  =  S  can  properly  apply 
to  it,  on  account  of  waves  which  produce  secondary  flexure  in 
the  plates  or  webs.  This  view  of  the  subject  is  one  quite  new 
and  indicates  that  our  present  knowledge  of  column  action  does 
not  apply  to  unusual  sections.  In  all  such  cases  there  is  only 
one  safe  guide,  namely,  tests  upon  the  actual  columns  or  upon 
models  which  represent  them  as  closely  as  possible. 

The  influence  of  the  weight  of  the  column  itself  has  not  been 
considered  in  the  preceding  pages.  When  the  column  is  in  a 
vertical  position,  it  might  appear  in  accordance  with  Art.  28, 
that  the  section  should  increase  continually  toward  the  base  in 
order  to  give  a  form  of  uniform  strength,  or  for  a  long  column> 


224  COLUMNS  OR  STRUTS  CHAP  IX 

where  lateral  flexure  is  to  be  feared,  that  the  greatest  section 
should  be  between  the  middle  of  the  length  and  the  base.  For 
usual  cases,  however,  the  necessary  increase  of  section  is  so  small 
as  to  be  inappreciable.  The  Doric  column  of  Greek  architec- 
ture had  its  greatest  diameter  at  the  base,  while  the  Ionic  column 
was  usually  of  constant  diameter  up  to  about  one-third  of  its 
height.  Considerations  of  beauty  rather  than  strength  governed, 
however,  the  evolution  of  these  ancient  forms. 

When  a  column  is  placed  in  a  horizontal  position,  its  weight 
causes  flexure  which  increases  the  deflection  and  stress  due  to 
the  direct  compression.  This  case  will  be  discussed  in  Art.  104, 
and  it  will  be  there  seen  that  the  direct  compression  is  sometimes 
applied  eccentrically  in  order  to  counteract  the  flexure  caused 
by  the  weight  of  the  column. 

Prob.  882.  Prove  that  4.49341  is  a  root  of  the  equation  /?  =  tan/9. 

Prob.  886.  Consult  the  engineering  periodicals  for  September,  1907, 
and  ascertain  facts  regarding  the  failure  of  the  great  cantilever  bridge 
over  the  St.  Lawrence  river  at  Quebec. 


ART.  89  PHENOMENA   OF  TORSION  225 


CHAPTER  X 

/ 

TORSION  OF  SHAFTS 
ART.  89.     PHENOMENA  or  TORSION 

When  applied  forces  cause  a  bar  to  twist  around  its  axis, 
ftorsional  stresses  '  arise.  If  a  rectangular  bar  has  one  end  fixed 
and  forces  are  applied  to  the  other  end  which  cause  twisting 
around  its  axis,  the  corners  of  the  bar  are  seen  to  assume  a  spiral 
form  similar  to  the  threads  of  a  screw.  By  experiments  like  those 
illustrated  in  Fig.  169c  it  has  been  proved  that  the  phenomena 
of  torsion  are  analogous  to  those  of  tension.  Under  small  twisting 
forces  the  deformation,  or  angle  of  twist,  is  proportional  to  the 
force,  so  that  the  bar  returns  to  its  original  form  on  the  removal 
of  that  force.  This  law  holds  until  an  elastic  limit  is  reached; 
beyond  this  limit  the  angle  of  twist  increases  more  rapidly  than 
the  force,  and  a  permanent  set  remains  when  the  force  is  removed. 
Under  further  increase  of  the  twisting  force,  the  deformation 
rapidly  increases  and  rupture  finally  occurs.  In  Fig.  169^  are 
seen  a  square  bar  and  a  round  bar  which  have  been  ruptured  by 
torsion. 

The  force  P  which  causes  the  twist  acts  in  a  direction  normal 
to  the  axis  of  the  bar  or  shaft  with  a  certain  lever  arm  p.  Fig.  89 
shows  a  horizontal  shaft  rigidly  fixed  at  one  end,  while  a  weight  P 
is  hung  on  a  lever  at  right  angles  to  the  axis  of  the  shaft.  Under 
the  twisting  moment  Pp  the  shaft  is  deformed,  so  that  an  originally 
straight  line  ab  becomes  the  helix  ad,  while  the  radial  line  cb  has 
moved  through  the  angle  bed.  The  angle  bed  is  evidently  propor- 
tional to  the  length  of  the  shaft,  while  the  angle  bad  is  independent 
of  that  length. 

The  product  Pp  is  the  moment  of  the  force  P  with  respect  to 
the  axis  of  the  shaft,  p  being  the  perpendicular  distance  from 
that  axis  to  the  line  of  direction  of  P,  and  is  called  the  'twisting 
moment '.  Whatever  be  the  number  of  forces  acting  upon  the 


226  TORSION  OF  SHAFTS  CHAP.  X 

shaft,  their  resulting  twisting  moment  may  always  be  represented 
by  a  single  product  Pp.  Thus,  if  the  forces  P\  and  P2  act 
with  lever  arms  pi  and  p2j  the  twisting  moment  Pipi+Pzpz  may 
be  caused  by  a  single  force  P  with  the  lever  arm  p. 

A  graphical  representation  of  the  phenomena  of  torsion  may 
be  made  in  the  same  manner  as  the  tension  diagram  of  Fig.  4, 
the  angles  of  torsion  being  taken  as  abscissas,  and  the  twisting 
moments  as  ordinates.  The  curve  is  then  a  straight  line  from 
the  origin  until  the  elastic  limit  of  the  material  is  reached,  when 
a  rapid  change  occurs  and  it  soon  becomes  nearly  parallel  to  the 
axis  of  abscissas.  The  total  angle  of  torsion,  like  the  total  ulti- 
mate elongation,  serves  to  compare  the  ductility  of  materials. 

The  principal  stress  which  occurs  in  torsion  is  that  of  shearing, 
each  section  of  the  bar  tending  to  shear  off  from  the  one  adjacent 

to  it.  The  direction  of 
the  shearing  stress  at 
any  point  in  the  sec- 
tion of  a  round  shaft 
is  normal  to  a  radius 
drawn  from  the  axis  to 
that  point,  and  the  sum 
of  the  moments  of  all 
Fig.  89  the  stresses  in  any  sec- 

tion is  equal  to  the  twisting  moment  Pp.  For  a  round  bar,  like 
that  of  Fig.  89,  a  radial  straight  line  cb  is  found  to  remain 
straight  when  displaced  to  the  position  cd,  provided  the  shear- 
ing elastic  limit  of  the  material  is  not  exceeded.  For  a  square  or 
rectangular  bar,  this  is  not  the  case  when  the  radial  line  is  drawn 
to  a  corner. 

Prob.  89a.  If  a  force  of  80  pounds  acting  at  18  inches  from  the  axis 
twists  a  shaft  15  degrees,  what  force  will  produce  the  same  result  when 
acting  at  4  feet  from  the  axis  ? 

Prob.  89&.  A  shaft  2  feet  long  is  twisted  through  an  angle  of  7 
degrees  by  a  force  of  200  pounds  acting  at  a  distance  of  6  inches  from 
the  axis.  Through  what  angle  will  a  shaft  of  the  same  size  and 
material  and  4  feet  long  be  twisted  by  a  force  of  500  pounds  acting  at 
a  distance  of  18  inches  from  the  axis? 


ART.  90 


THE  TORSION  FORMULA 


227 


ART.  90.    THE  TORSION  FORMULA 

It  has  been  found  by  experiment  that  the  laws  governing  the 
stresses  in  a  section  of  a  round  bar  under  torsion  are  similar  to 
those  stated  for  beams  in  Art.  40,  provided  the  elastic  limit  is 
not  exceeded.  The  shearing  unit-stresses  are  proportional  to  their 
distances  from  the  axis,  because  it  is  observed  that  any  radius, 
such  as  cb  in  Fig.  89,  remains  a  straight  line  when  it  is  displaced 
by  the  twisting  into  the  position  cd.  The  law  of  static  equilibrium 
requires  that  the  sum  of  the  moments  of  these  shearing  stresses 
shall  be  equal  to  the  twisting  moment,  or, 

Resisting  Moment = Twisting  Moment 
and  this  condition  will  now  be  expressed  in  algebraic  language. 


Fig.  90a 

Let  P  be  the  force  acting  at  a  distance  p  from  the  axis  about 
which  the  twisting  takes  place,  then  the  value  of  the  twisting 
moment  is  Pp.  To  find  the  resisting  moment,  let  c  be  the  distance 
from  the  axis  to  the  outside  of  the  circular  cross-section  where  the 
shearing  unit-stress  is  S.  Then,  since  the  shearing  unit-stresses 
vary  as  their  distances  from  the  axis  of  the  shaft, 

S/c  =  unit-stress  at  a  unit's  distance  from  axis 
5  .  z/c  =  unit-stress  at  a  distance  z  from  axis 
da  .  Sz/c  =  total  stress  on  an  elementary  area  da 
da  .  Sz2/c  =  moment  of  this  stress  with  respect  to  axis 
2  da  .  Sz2/c  =  internal  resisting  moment 

Since  S  and  c  are  constants,  this  resisting  moment  may  be 
written  (S/c)  Ida  .  z2.  But  the  quantity  Ida  .  z2,  being  the  sum 
of  the  products  obtained  by  multiplying  each  element  of  area  by 
the  square  of  its  distance  from  the  axis,  is  the  polar  moment  of 


228  TORSION  OF  SHAFTS  CHAP,  x 

inertia  of  the  cross-section  and  may  be  denoted  by  /.  The  resisting 
moment  of  the  internal  stresses  hence  is  5  .  J/c,  and  equating  this 
to  the  twisting  moment,  there  results, 

S,J/c  =  Pp  or  S=Pp.c/J  (90) 

which  is  the  fundamental  formula  for  investigating  round  bars 
and  shafts  that  are  subject  to  torsion.  It  will  be  called  the  torsion 
formula,  but  it  must  be  borne  in  mind  that  it  does  not  apply 
to  square  or  rectangular  sections;  these  will  be  discussed  in 
Art.  99. 

The  flexure  formula  S  .  I/c=M  for  beams  has  a  close  analogy 
with  the  torsion  formula.  In  the  flexure  formula,  c  is  the  distance 
from  the  neutral  axis  to  the  remotest  fiber  and  that  axis  lies  in 
the  plane  of  the  cross-section;  in  the  torsion  formula,  c  is  the 
radius  of  the  outside  circumference  of  the  round  bar.  In  the 
flexure  formula,  S  is  a  tensile  or  compressive  unit-stress  which 
is  normal  to  the  section  area ;  in  the  torsion  formula,  S  is  a  shearing 
unit-stress  which  acts  along  the  section  and  normal  to  the 
radius.  In  the  flexure  formula,  M  is  the  bending  moment  oi 
the  external  forces;  in  the  torsion  formula,  Pp  the  twisting 
moment  of  the  external  forces. 

By  the  help  of  the  torsion  formula,  three  problems  like  those 
of  Arts.  48,  49,  50,  may  be  discussed  for  round  bars  or  shafts. 
When  the  dimensions  and  the  allowable  unit-stress  are  given, 
it  may  be  used  to  compute  the  safe  twisting  moment  Pp.  When 
the  unit-stress  5S  and  the  moment  Pp  are  given,  it  may  be  used 
to  design  a  shaft  or  bar,  by  determining  dimensions  which  will 
give  a  value  for  J/c  equal  to  Pp/S.  The  results  obtained  in  this 
discussion  are  only  valid  when  the  shearing  unit-stress  S  does  not 
exceed  the  elastic  limit  of  the  material,  but  it  is  shown  in  Art.  94 
how  the  formula  may  be  used  for  cases  of  rupture. 

In  the  discussion  of  shafts,  the  moments  of  inertia  of  cross- 
sections  are  required  with  respect  to  a  point  at  the  center  of  the 
shaft  and  not  with  respect  to  an  axis  in  the  same  plane,  as  in  beams 
and  columns.  The  'polar  moment  of  inertia  '  of  a  surface  is 
defined  as  the  sum  of  the  products  obtained  by  multiplying  each 
elementary  area  by  the  square  of  its  distance  from  the  center  of 


ART.  90  THE  TORSION  FORMULA  229 

gravity  of  the  surface.  Thus  if  da  is  any  elementary  area  and 
z  its  distance  from  the  center,  the  quantity  Ida  .  z2  is  the  polar 
moment  of  inertia  of  the  surface. 

In  the  figure  let  da  be  any  elementary  area  and  y  its  distance 
from  an  axis  AB  passing  through  the  center  of  gravity  of  the 
section ;  then  Ida  .  y2  is  the  moment  of 
inertia  with  respect  to  this  axis  AB  (Art.  43). 
Also,  if  oc  is  the  distance  from  da  to  an  axis 
CD  which  is  normal  to  AB}  then  Ida  .  x2 
is  the  moment  of  inertia  with  respect  to  CD. 
But,  since  z2  =  x2  +  y2,  the  product  Ida.z2 
is  equal  to  Ida  .  x2  -f  Ida  .  y2 ;  that  is, 
the  polar  moment  of  inertia  is  equal  to  the  sum  of  the  moments 
of  inertia  taken  with  respect  to  any  two  rectangular  axes. 

By  the  aid  of  the  above  principle,  the  polar  moment  of  inertia 
/  is  readily  found  for  a  solid  or  hollow  circular  section  from  the 
values  of  /  given  in  Art.  43.  Let  d  be  the  diameter  of  a  solid 
section  or  the  outside  diameter  of  a  hollow  one,  and  let  d\  be  the 
inside  diameter  of  a  hollow  one;  then, 

for  a  solid  shaft,          c  =  %d,        J 

for  a  hollow  shaft,  c=%d,  J 
The  polar  radius  of  gyration  r,  defined  by  the  equation  /  =ar2> 
where  a  is  the  section  area,  is  also  sometimes  used  in  formulas; 
it  is  the  radius  of  a  circumference  along  which  the  entire  area 
might  be  concentrated  and  have  the  same  polar  moment  of  inertia 
as  the  actual  distributed  area.  The  value  of  r  is  always  less  than 
%d;  for  a  solid  circular  section,  r2  =  %d2;  for  a  hollow  circular  sec- 
tion, r2  =  J(^2  +  ^i2)- 

Prob.  90a.  Three  forces  of  70,  90,  and  120  pounds  act  at  distances 
of  8,  u,  and  6  inches  respectively  from  the  axis,  and  at  different  dis- 
tances from  the  end  of  a  shaft,  the  direction  of  rotation  of  the  second 
force  being  opposite  to  that  of  the  others.  Find  the  three  values  of 
the  twisting  moment  Pp. 

Prob.  906.  A  circular  shaft  is  subjected  to  a  maximum  shearing 
unit-stress  of  2  ooo  pounds  when  twisted  by  a  force  of  90  pounds  at  a 
distance  of  27  inches  from  the  axis.  Compute  the  ur  it-stress  pro- 
duced by  a  force  of  40  pounds  at  21  inches  from  the  axis. 


230  TORSION  OF  SHAFTS  CHAP.  X 


ART.  91.    TRANSMITTING  POWER  BY  SHAFTS 

Power  from  a  motor  is  often  transmitted  to  a  shaft  through  a 
belt,  and  then  the  shaft  transmits  the  power  to  the  places  where 
the  work  is  to  be  performed.  The  shaft  of  a  turbine  wheel 
transmits  the  power  generated  by  water  in  passing  through  the 
wheel  directly  to  dynamos  or  other  machinery.  The  shaft  of 
an  ocean  steamer  transmits  the  power  of  the  engines  directly  to 
the  screw  propellers.  In  all  these  cases  the  shaft  is  subject  to 
a  twisting  moment  Pp  which  produces  in  it  shearing  stresses 
and  causes  it  to  have  a  certain  angle  of  twist. 

The  twisting  moment  Pp  due  to  the  transmission  of  H  horse- 
powers through  a  shaft  may  be  found  as  follows:  Suppose  P 
to  be  the  tangential  force  brought  by  a  belt  on  the  circum- 
ference of  a  pulley  of  radius  pi  and  let  n  be  the  number  of  revolu- 
tions made  by  the  shaft  and  pulley  in  one  minute.  In  one  revo- 
lution the  force  P  overcomes  resistance  through  the  distance  2np 
and  the  work  PX2np  or  27iPp  is  transmitted  through  the  shaft; 
in  n  revolutions  the  work  2nnPp  is  transmitted.  Now  let  N 
be  the  number  of  units  of  work  per  minute  which  constitute  one 
horse-power;  then  2nnPp  =  NH)  or  Pp  =  NH/2xn.  In  the 
English  system  of  measures,  P  is  in  pounds,  p  in  inches,  and  N  is 
396  ooo  pound-inches  per  minute.  In  the  metric  system  of 
measures,  P  is  m,kilograms,  p  in  centimeters,  and  N  is  450  ooo 
kilogram-centimeters  f)erfcminut£.  Accordingly, 

Pp  =  6soT>oH/n  or  Pp  =  ^i  62oH/n  (91) 

the  first  being  for  the  English  and  the  second  for  the  metric 
system  of  measures,  while  Pp=  (N/2n)H/n  applies  to  all  systems. 

The  above  formula  shows  that  the  twisting  moment  Pp  varies 
directly  as  the  transmitted  horse-power  and  inversely  as  the  speed 
of  revolution.  Therefore,  when  the  speed  is  low,  as  it  is  in  starting, 
the  full  power  should  not  be  applied  to  a  shaft,  for  it  might  render 
the  twisting  moment  so  great  as  to  injure  the  material  or  to  cause 
rupture. 

Metallic  shafts  are  usually  round  and  the  word  '  shaft ',  when 


ART.  92  SOLID    AND    HOLLOW    SHAFTS  231 

used  without  qualification,  means  a  solid  round  cylinder,  properly 
supported  in  bearings.  Square  shafts  are  rarely  used  except 
for  wooden  water  wheels,  and  the  torsion  formula  (90)  does  not 
apply  to  these  unless  modified  in  the  manner  indicated  in  Art.  99. 
Hollow  sections  have  been  much  used  since  1900  for  the  large 
shafts  of  ocean  steamers.  These  are  advantageous  in  being  of 
lighter  weight  than  solid  shafts  of  the  same  strength  and  capacity, 
as  the  investigation  in  Art.  95  will  show;  these  shafts  are  forged 
upon  a  mandrel  and  the  inside  surface  is  hence  subject  to  the 
same  treatment  as  the  outside  surface. 

In  designing  a  shaft  to  resist  a  given  twisting  moment  Pp 
the  factors  of  safety  may  be  based  upon  the  average  shearing 
strengths  of  materials  which  are  given  in  Art.  6.  The  rule  that 
an  allowable  unit-stress  S  should  not  exceed  the  shearing  elastic 
limit  of  the  material  should  also  be  observed,  although  there  is 
considerable  uncertainty  as  to  the  average  values  of  this  limit. 
Probably  the  elastic  limits  for  shearing  are  about  three-fourths 
of  those  for  tension. 

Prob.  910.  Show  that  the  metric  horse-power  is  1.4  percent  less 
than  the  English  horse-power;  also  that  one  kilogram-meter  is  equiva- 
lent to  7.233  foot-pounds. 

Prob.  Qlb.  Find  the  horse-power  that  can  be  transmitted  by  a  cast- 
iron  shaft  3  inches  in  diameter  when  making  10  revolutions  per  min- 
ute, the  value  of  S  not  to  exceed  i  200  pounds  per  square  inch. 


ART.  92.    SOLID  AND  HOLLOW  SHAFTS 

For  solid  round  shafts  of  diameter  d,  the  values  of  J  and  c 
are  -jV^4  and  %d,  and  the  torsion  formula  (90)  reduces  to 

Pp=^nd3S  or  S=i6Pp/nd3 

which  may  be  used  for  the  discussion  of  shafts  when  the  twisting 
moment  Pp  is  required  or  given.  The  usual  case,  however,  is 
that  of  the  transmission  of  power,  where  the  value  of  Pp  for  the 
English  system  of  measures  is  63  o^oH/n,  as  shown  in  Art.  91 ; 
inserting  this  in  the  last  formula,  it  reduces  to  the  forms, 

5=321  oooH/nd3  or  d=6S.s(H/nS^  (92) 


232  TORSION  OF  SHAFTS  CHAP,  x 

The  first  of  these  may  be  used  for  investigating  the  strength  of  a 
given  shaft  when  transmitting  H  horse-powers  and  making 
n  revolutions  per  minute.  The  second  may  be  used  to  determine 
the  diameter  of  a  shaft  to  transmit  H  horse-powers  with  n  revo- 
lutions per  minute,  the  allowable  unit-stress  for  the  circumfer- 
ence of  the  shaft  being  S.  In  both  equations  d  must  be  taken 
in  inches  and  5  in  pounds  per  square  inch. 

These  equations  show  that  the  shearing  unit-stress  5  at  the 
circumference  of  a  solid  shaft  varies  directly  as  the  transmitted 
power,  inversely  as  the  speed  of  the  shaft,  and  inversely  as  the 
cube  of  its  diameter.  Hence  for  a  given  S  and  n,  the  diameter 
d  changes  slowly  with  H\  if  H  is  doubled,  d  is  increased  only 
26  percent,  and  the  section  area  is  increased  59  percent. 

For  example,  let  it  be  required  to  design  a  solid  wrought- 
iron  shaft  to  transmit  90  horse-power  when  making  250  revolu- 
tions per  minute.  Here  the  factor  of  safely  may  be  about  6, 
or  S  may  be  about  7  ooo  pounds  per  square  inch.  Then  from 
the  above  formula  the  diameter  d  is  found  to  be  2§  inches. 

Hollow  forged  steel  shafts  are  much  used  for  ocean  steamers, 
since  their  strength  is  greater  than  solid  shafts  of  the  same  section 
area.  Let  di  be  the  outside  and  d2  the  inside  diameter;  then  the 
value  of  J  is  aVK^i4"  d24)  and  that  of  c  is  ^d\.  Inserting  these 
in  the  torsion  fonnula  and  also  the  value  of  Pp  from  (91),  it  be- 
comes, for  English  measures, 

S(di*  -  d2*)/di  =  321  oooH/n 
which  is  the  equation  for  use  in  investigation  and  design. 

For  example,  a  nickel -steel  shaft  of  17  inches  outside  diameter 
is  to  transmit  16000  horse-powers  at  50  revolutions  per  minute; 
let  it  be  required  to  find  the  inside  diameter  so  that  the  unit- 
stress  5  may  be  25  ooo  pounds  per  square  inch.  Here  every- 
thing is  given  except  d2,  and  from  the  equation  its  value  is  found 
to  be  ii  inches  nearly.  The  area  of  the  cross-section  of  this 
shaft  will  be  about  132  square  inches,  and  its  weight  per  linear 
foot  about  449  pounds.  A  solid  shaft  having  the  same  strength 
will  require  a  diameter  of  16  inches,  its  cross-section  will  be 
201  square  inches,  and  its  weight  per  linear  foot  about  683  pounds. 


ART.  93  TWIST   OF    SHAFTS  233 

The  reason  why  economy  is  promoted  by  the  use  of  a  hollow 
shaft  is  the  same  as  that  given  in  Art.  51  for  a  hollow  beam; 
namely,  that  material  is  removed  from  the  axis  where  it  is  but 
little  stressed  and  placed  farther  away  where  it  is  efficient  in 
resisting  the  twisting  moment.  By  comparing  the  above  formulas 
for  solid  and  hollow  shafts,  it  is  seen  that  they  become  the  same 
when  d*  equals  (di4-d24)/%.  When  the  section  areas  of  the 
solid  and  hollow  shafts  are  equal,  d2  must  equal  di2~d22.  From 
these  two  equations,  it  follows  that  the  ratio  of  the  strength 
of  a  hollow  shaft  to  that  of  a  solid  one  of  the  same  section  area 
is  (2di2  —  d2)/ddi.  For  example,  let  d  =  i2  inches  and  ^1=20 
inches;  then  d2  is  16  inches  from  the  condition  of  equal  section 
areas,  and  the  ratio  of  strengths  is  2.73,  that  is,  the  strength  of 
the  hollow  shaft  is  173  percent  greater  than  the  solid,  one. 

Prob.  92a.  Find  the  factor  of  safety  for  a  wrought-iron  shaft  2j 
inches  in  diameter  when  transmitting  25  horse-power  while  making  TOO 
revolutions  per  minute. 

Prob.  926.  If  a  hollow  shaft  has  the  same  section  area  as  a  solid 
one,  the  inside  diameter  being  one-half  the  outside  diameter,  find  the 
ratio  of  the  strength  of  the  hollow  shaft  to  that  of  the  solid  one. 


ART.  93.    TWIST  OF  SHAFTS 

The  angle  of  twist  through  which  a  radius  on  one  end  of  a 
round  shaft  moves  under  an  applied  twisting  moment  can  be 
determined,  when  the  unit-stress  5  does  not  exceed  the  elastic 
limit,  from  the  definitions  in  Art.  15  with  the  help  of  the  torsion 
formula  (90).  Let  the  section  be  circular,  J  its  polar  moment 
of  inertia  with  respect  to  the  axis  about  which  rotation  occurs, 
c  the  distance  from  that  axis  to  the  remotest  part  of  the  section 
where  the  shearing  unit-stress  is  S,  and  F  the  shearing  modulus 
of  elasticity  of  the  material.  The  modulus  F  is  defined  by  S/e, 
where  £  is  the  deformation  per  unit  of  length.  Let  /  be  the  length 
of  the  shaft,  and  <f>  the  angle  expressed  in  radians  through  which 
one  end  is  twisted  with  respect  to  the  other,  that  is  the  angle 
deb  in  Fig.  89a.  In  this  figure  the  arc  bd  is  the  deformation  in 
the  length  I]  this  deformation  is  c<j)  and  hence  the  unit-deforma- 


234  TORSION  OF  SHAFTS  CHAP.  X 

tion  is  c<j)/l.  Accordingly  F  =  Sl/c</>,  and  replacing  S  by  its 
value  Pp  .  c/J  there  is  obtained, 

F=Pp.l/J<f>  or  </>=Pp.l/FJ  (93) 

The  first  of  these  equations  may  be  used  to  compute  values  of 
the  shearing  modulus  F  from  observed  values  of  </>,  while  the 
second  is  for  the  determination  of  0  when  F  is  given. 

In  order  to  compute  the  shearing  modulus  of  elasticity  from 
an  observed  value  of  the  angle  of  twist  the  arc  0  should  be  replaced 
by  7r<£o/i8o,  where  00  is  the  angle  in  degrees;  then, 

<h  =  V'ZPpl/FJ  or  F  =  ^.zPpl/J^Q 

By  taking  several  corresponding  values  of  Pp  and  0o  within  the 
elastic  limit,  a  good  determination  of  F  can  be  made.  For 
example,  the  cast-iron  specimen  shown  in  Fig.  169c  was  10  inches 
long  and  0.83  inches  in  diameter,  and  it  twisted  through  an 
angle  of  1.3  degrees  under  a  twisting  moment  of  600  inch-pounds; 
here  7=0.0466  inches4,  and  the  formula  gives  ^  =  5670000 
pounds  per  square  inch.  In  this  manner  the  average  values  of 
F  have  been  found  to  be  about  6  ooo  ooo  pounds  per  square 
inch  for  cast  iron  and  about  12  ooo  ooo  pounds  per  square  inch 
for  steel. 

The  angle  of  twist  which  occurs  when  a  shaft  is  under  stress 
in  the  transmission  of  power  may  be  determined  in  a  similar 
manner.  Let  H  horse-power  be  transmitted  when  the  shaft 
makes  n  revolutions  per  minute ;  then  Pp  =  63  o^oH/n  in  English 
measures  (Art.  91).  Let  <j6o  be  angle  of  twist  in  degrees;  then 
(f>  is  to  be  replaced  by  7r00/i8o.  The  formula  now  becomes, 

<po  =  3  610  oooHl/nFJ  (93)' 

in  which  /  must  be  taken  in  inches,  J  in  inches4,  and  F  in  pounds 
per  square  inch.  For  example,  let  a  steel  shaft  125  feet  long, 
17  inches  outside  diameter,  and  n  inches  inside  diameter  trans- 
mit 16  ooo  horse-powers  at  50  revolutions  per  minute.  Here 
H  =  16000  horse-powers,  l  =  i  500  inches,  ^  =  50,  F  =  i2  ooo  ooo 
pounds  per  square  inch,  /  =  6  765  inches4.  Then  from  the 
formula  (93) '  there  is  found  (£0  =  21.4  degrees,  which  is  the  angle 
through  which  a  point  on  one  end  is  twisted  relative  to  the  corre- 


ART.  94  RUPTURE   BY   TORSION  235 

spending  point  on  the  other  end.  The  angle  through  which 
a  straight  line  on  the  outside  of  the  shaft  is  twisted,  that  is,. the 
angle  bad  in  Fig.  89,  is  C(f>Q/l  =  &.$X2 1.4/1  500  =  0.12  degrees. 

The  above  formulas  show  that  the  angle  of  twist  is  propor- 
tional to  the  length  of  the  shaft,  and  this  is  found  to  be  also  the 
case  after  the  elastic  limit  of  the  material  is  exceeded,  but  the 
angle  is  then  very  much  greater  than  that  given  by  the  formulas, 
In  the  case  of  the  round  cast-iron  bar  shown  in  Fig.  169c,  the 
angle  of  twist  at  rupture  was  12.3  degrees;  this  bar  had  a  length 
of  10  inches  and  a  diameter  of  0.83  inches,  and  it  broke  under  a 
twisting  moment  of  3  910  pound-inches.  For  the  square  steel 
bar  in  Fig.  169c,  which  was  12  inches  long,  the  angle  of  twist  at 
rupture  was  about  900  degrees;  further  facts  regarding  this  square 
bar  are  given  in  Art.  99. 

Prob.  93a.  A  solid  steel  shaft  125  feet  long  and  16  inches  in 
diameter  transmits  8  ooo  horse-powers  at  a  speed  of  25  revolutions 
per  minute.  Compute  the  angle  of  twist. 

Prob.  936.  A  wrought-iron  shaft  5  feet  long  and  2  inches  in  diame- 
ter is  twisted  through  an  angle  of  7  degrees  by  a  force  of  5  ooo  pounds 
acting  at  6  inches  from  its  axis,  and  on  the  removal  of  the  force  it 
springs  back  to  its  original  position.  Compute  the  shearing  modulus 
of  elasticity. 

ART.  94.    RUPTURE  BY  TORSION 

When  a  round  bar  or  shaft  is  twisted  to  the  point  of  rupture, 
failure  occurs  first  at  the  outside  circumference,  and  this  is 
rapidly  propagated  inwards  until  complete  shearing  is  effected. 
The  torsion  formula  S.J/c  =  Pp,  does  not,  however,  hold  for 
cases  for  rupture,  so  that  a  value  of  S  computed  from  it  for  the 
data  of  failure  does  not  closely  agree  with  the  ultimate  shearing 
strength  of  the  material. 

A  value  of  the- unit-stress  S  computed  by  the  torsion  formula 
from  a  twisting  moment  Pp  which  causes  rupture,  may  be  called 
the  '  computed  twisting  strength .'  of  the  material,  in  analogy  with 
the  computed  flexural  strength  obtained  for  the  rupture  of  a 
beam  by  the  use  of  the  flexure  formula  (Art.  52).  The  following 


236  TORSION  OF  SHAFTS  CHAP.  X 

are  approximate  average  values  of  the  computed  ultimate  -twisting 

strength  of  different  materials: 

Timber  Su=   2  ooo  pounds  per  square  inch 

Cast  Iron  Su  =  3o  ooo  pounds  per  square  inch 

Wrought  Iron  SU  =  S5  ooo  pounds  per  square  inch 
Structural  Steel  Su  =  6s  ooo  pounds  per  square  inch 
Strong  Steel  ,$^  =  90  ooo  pounds  per  square  inch 

By  the  use  of  these  average  values,  computations  may  be  made 
on  the  rupture  of  round  shafts,  that  is,  the  probable  twisting 
moment  which  will  cause  a  given  shaft  to  rupture,  or  the  probable 
size  of  a  shaft  which  will  fail  under  a  given  twisting  moment, 
may  be  roughly  ascertained. 

By  comparing  the  above  values  with  the  ultimate  shearing 
strengths  given  in  Art.  6,  it  will  be  seen  that  they  are  all  larger 
with  the  exception  of  timber.  The  statement  is  sometimes  made 
that  it  might  be  anticipated  that  the  two  sets  of  values  should 
be  the  same,  but  this  rests  on  no  reasonable  basis.  The  ultimate 
shearing  strength  of  a  material  is  a  physical  constant,  but  the 
values  of  Su  have  no  physical  meaning,  for  they  have  been  com- 
puted from  a  formula  which  does  not  correctly  represent  the 
distribution  of  the  internal  stresses  for  cases  of  rupture.  The 
quantity  Su  is  usually  called  the  'modulus  of  rupture  for  torsion  ', 
but  it  seems  wise  to  abandon  this  term  and  to  use  one  which 
gives  no  erroneous  impression. 

When  that  twisting  moment  Pp  is  reached  in  a  test  where  the 
angles  of  twist  (f>  no  longer  increase  uniformly,  the  shearing 
elastic  limit  of  the  material  is  reached,  and  the  elastic  limit  for 
shearing  may  be  correctly  computed  from  the  torsion  formula 
S  =  Pp.c/J.  The  shearing  elastic  limit  of  metals  is  always 
less  than  those  for  compression,  and  in  general  may  be  taken  at 
about  one-third  of  the  computed  twisting  strength  Su.  Experi- 
ments on  timber,  brick,  and  stone  have  been  so  few  in  number, 
that  it  is  difficult  to  state  average  values  of  their  shearing  elastic 
limits. 

In  Fig.  188  is  shown  a  cast-iron  bar,  0.83  inches  in  diameter 
which  broke  by  torsion  under  a  twisting  moment  of  3  910  inch- 


ART.  95  STRENGTH    AND    STIFFNESS  237 

pounds.  The  computed  ultimate  twisting  strength  for  this  case 
is  5  =  34800  pounds  per  square  inch,  which  is  higher  than  the 
average  for  cast  iron.  The  square  bar  of  medium  steel  shown 
in  Fig.  169c,  which  was  0.75  inches  square,  broke  under  a  twisting 
moment  of  5  800  inch-pounds;  the  torsion  formula  (90)  cannot 
be  applied  to  this  case,  as  the  discussion  in  Art.  99  will  show. 

Prob.  94.  Compute  the  probable  diameter  of  a  wrought-iron  shaft 
which  will  rupture  by  torsion  when  twisted  by  a  force  of  47  pounds 
acting  at  a  distance  of  47  inches  from  the  axis. 

ART.  95.     STRENGTH  AND  STIFFNESS 

The  strength  of  a  shaft  or  of  a  bar  under  torsion  is  measured 
by  the  twisting  moment  that  it  can  carry.  Hence  the  strengths 
of  different  shafts  vary  as  their  values  of  S  .  J/c;  or  for  the  same 
material,  the  strengths  vary  as  the  values  of  J/c.  For  example, 
let  there  be  two  solid  shafts  of  the  same  material,  the  diameter 
of  the  first  being  d  and  that  of  the  second  2^;  since  the  section 
factor  J/c  is  eight  times  as  large  for  the  second  as  for  the  first, 
it  follows  that  their  strengths  arc  in  the  same  ratio.  Thus  the 
strengths  of  solid  shafts  of  the  same  material  vary  as  the  cubes 
of  their  diameters. 

A  general  comparison  of  the  strength  of  a  round  hollow  shaft 
with  that  of  a  solid  one  having  the  same  section  area  a  will  now 
be  made.  Let  d  be  the  diameter  of  the  solid  shaft,  di  the  outside 
and  &2  the  inside  diameter  of  the  hollow  shaft  ;  then  d2  =  di2  —  d22 
is  the  condition  that  the  section  areas  shall  be  equal.  Now, 
for  the  hollow  shaft, 


and  in  the  same  manner  for  the  solid  shaft, 


Therefore,  dividing  the  first  by  the  second,  and  letting  a  denote 
the  ratio  di/d2l  there  is  found 

hollow/solid-  (a2+  i)/a(a2-i)* 

which  is  the  ratio  of  the  strength  of  a  hollow  shaft  to  that  of  a 
solid  one  of  the  same  section  area.     When  di  is  double  d2)  the 


238  TORSION  OF  SHAFTS  CHAP,  x 

value  of  a  is  2,  and  the  hollow  shaft  is  about  44  percent  stronger 
than  the  solid  one. 

The  stiffness  of  a  shaft  under  torsion  is  measured  by  the 
twisting  moment  it  can  carry  with  a  given  angle  of  twist.  By 
Art.  93  it  is  seen  that  Pp  varies  directly  with  /,  and  hence  the 
stiffness  of  a  shaft  varies  directly  as  its  polar  moment  of  inertia. 
For  the  hollow  shaft, 


and  for  the  solid  shaft  of  equal  section  area, 


Therefore,  dividing  the  first  by  the  second,  and  designating  the 
ratio  di/dz  by  a,  there  is  found 

hollow/solid  =  (a2  +  i)/(a2  -  1) 

which  is  the  ratio  of  the  stiffness  of  a  hollow  shaft  to  that  of  a 
solid  one  of  the  same  section  area.  When  d\  is  double  d2,  the 
value  of  a  is  2,  and  the  hollow  shaft  is  about  67  percent  stiffer 
than  the  solid  one. 

Shafts  of  the  same  material  and  length  are  of  the  same  strength 
when  their  values  of  J/c  are  equal,  and  of  the  same  stiffness  when 
their  values  of  /  are  equal.  It  is  easy  to  show  that  the  per- 
centage of  weight  saved  by  using  a  hollow  shaft  instead  of 
a  solid  one  is, 

for  equal  strength          100  —  ioo[(a4—  a2)  /(a2  +  1)2]* 
for  equal  stiffness  100  —  ioo[(o:2  —  i)/(a2+i)]* 

in  which  a  is  the  ratio  d\/d2  which  will  be  determined  by  practical 
considerations  concerning  ease  of  manufacture  and  operation. 
When  a  =2,  the  percentages  saved  are  21.7  for  equal  strength 
and  22.5  for  equal  stiffness. 

Prob.  95.  Compare  the  strengths  of  two  shafts  when  stressed  to 
their  elastic  limits;  the  first  shaft  is  solid,  21  inches  in  diameter,  and 
has  an  elastic  limit  of  25000  pounds  per  square  inch;  the  second 
shaft  is  hollow,  18  inches  outside  and  9  inches  inside  diameter,  and 
its  elastic  limit  is  45  ooo  pounds  per  square  inch.  If  the  price  per 
pound  of  the  first  shaft  is  18}  cents,  what  price  per  pound  could  one 
afford  to  pay  for  the  second  shaft  ? 


ART.  96  SHAFT    COUPLINGS  239 


ART.  96.    SHAFT  COUPLINGS 

At  A  and  B  in  Fig.  96  are  shown  the  end  and  side  views  of 
a  flange  coupling  for  a  shaft,  the  flanges  being  connected  by 
bolts.  These  bolts,  in  transmitting  the  torsion  from  one  flange 
to  the  other,  are  subject  to  shearing  stress,  and  they  must  be 
of  sufficient  strength  to  safely  carry  it.  This  shear  does  not 
differ  in  character  from  that  in  the  main  body  of  the  shaft,  and 
it  is  the  greatest  upon  the  side  of  the  bolt  most  remote  from  the 
axis  of  the  shaft. 


Let  J  be  the  polar  moment  of  inertia  of  the  cross-section  of 
a  solid  shaft,  c  its  radius,  and  S  the  shearing  unit-stress  on  the 
outer  surface.  Let  /i  be  the  polar  moment  of  inertia  of  the 
cross-section  of  the  bolts  with  respect  to  the  axis  of  the  shaft,  c\ 
the  distance  from  the  axis  of  the  shaft  to  the  side  of  the  bolts 
farthest  from  the  axis,  and  let  the  shearing  unit-stress  on  that 
side  be  required  to  be  the  same  as  that  on  the  outer  surface  of 
the  shaft.  Then,  in  order  that  the  bolts  may  be  equal  in  strength 
to  the  shaft,  it  is  necessary  that  J/c  should  equal  J\/c\.  The 
polar  moment  of  inertia  of  the  section  of  one  bolt  with  respect 
to  the  axis  of  the  shaft  is  equal  to  its  polar  moment  with  respect 
to  its  own  axis  plus  the  section  area  of  the  bolt  into  the  square 
of  the  distance  between  the  two  axes. 

Let  D  be  the  diameter  of  the  shaft,  d  the  diameter  of  each 
of  the  bolts,  h  the  distance  of  the  center  of  a  bolt  from  the  axis 
o¥  the  shaft,  and  n  the  number  of  bolts;  then 


and  by  equating  these  values  there  is  found 

D\d+2h)=nd2(d2+Sh2) 
which  is  the  necessary  relation  between  the  quantities  in  order 


240  TORSION  OF  SHAFTS  CHAP,  x 

that  the  bolts  may  be  equal  in  strength  to  the  shaft,  provided 
the  material  be  the  same.  From  this  formula  the  number  of  bolts 
required  is  easily  found  when  d,  h,  and  D  are  given. 

This  formula  is  an  awkward  one  for  determining  d,  and  hence 
it  is  often  assumed  that  the  shear  is  uniformly  distributed  over 
the  bolts,  or  that  c\  =  h  and  J\  =  \nd2h2.  This  amounts  to  the 
same  thing  as  regarding  d  as  small  compared  to  h,  and  the  expres- 
sion then  reduces  to 

D3=4nhd2  or  d  =  %(D3/nh)* 

from  which  an  approximate  value  of  d  can  be  found  when  the 
number  of  bolts  is  given. 

The  above,  supposes  the  shaft  to  be  solid.  When  it  is  hollow, 
with  outside  diameter  D  and  inside  diameter  d\,  the  D3  in  the 
above  expressions  is  to  be  replaced  by  D3  —  di*/D. 

The  case  shown  at  CD  in  Fig.  96  is  one  that  would  not  occur 
in  practice,  but  it  is  here  introduced  in  order  to  indicate  that 
the  bolts  would  be  subject  to  a  flexural  as  well  as  a  shearing 
stress.  It  is  clear  that  the  flexural  stress  will  increase  with  the 
length  of  the  bolts,  and  that  they  should  be  greater  in  diameter 
than  for  the  case  of  pure  shearing.  The-  flexural  stress  will  also 
depend  upon  the  work  transmitted  by  the  shaft.  This  case  will 
be  investigated  in  Art.  98  in  connection  with  the  discussion  of 
the  pin  of  a  crank  shaft. 

Prob.  96a.  A  solid  shaft  6  inches  in  diameter  is  coupled  by  bolts 
ij  inches  in  diameter  with  their  centers  5  inches  from  the  axis.  How 
many  bolts  are  necessary  ? 

Prob.  96&.  A  hollow  shaft  17  inches  in  outside  and  n  inches  in 
inside  diameter  is  to  be  coupled  by  12  bolts  placed  with  their  centers 
20  inches  from  the  axis.  What  should  be  the  diameter  of  the  bolts  ? 

ART.  97.    A  SHAFT  WITH  CRANK 

A  crank  pin,  CD  in  Fig.  97,  is  subject  to  a  pressure  W  from 
the  connecting  rod  of  the  steam  engine,  this  pressure  being  uni- 
formly distributed  over  nearly  the  length  of  the  pin.  The  pressure 
varies  at  different  positions  in  the  stroke  of  the  engine,  but  for 


ART.  97  A   SHAFT  WITH    CRANK  241 

ordinary  computations  may  be  taken  at  from  10  to  20  percent 
greater  than  the  total  mean  pressure  of  the  steam  on  the  piston; 
to  this  may  be  added  the  weight  of  the  connecting  and  piston 
rods  in  case  these  should  be  vertical  in  position. 

The  maximum  pressure  W  causes  a  cross-shear  in  the  crank 
pin  at  the  section  C,  and  it  also  causes  a  flexural  stress  at  the 
end  C  due  to  a  uniform  load  over  the  cantilever  CD.  These 
stresses  may  be  computed  by  the  shear  and  flexure  formulas  of 
Art.  41.  The  compressive  or  bearing  stress  upon  the  pin  is 
usually  also  to  be  considered,  this  being  estimated  per  square 
unit  of  diametral  area  in  the  same  manner  as  the  sidewise  pressure 
on  a  rivet  (Art.  33.) 

In  the  figure  the  pressure  W  is  shown  acting  parallel  to  the 
crank  arm  BC,  but  it  acts  at  all  angles  to  this  arm  during  one 
revolution  of  the  shaft  AB.  When  it  acts 
at  right  angles,  the  crank  arm  is  a  can-  (~~^~ 

tilever  beam,    and    the  maximum   flexural     ^— - 

stresses  which  occur  at  B  may  be  computed 

by  the  flexure  formula  (41).     The  flexural 

unit-stresses    in  both    pin  and  crank    arm 

alternate  from  tension  to  compression  as  the 

arm  revolves,  for  the  pressure  W  comes  on  Yig.  97 

opposite  sides  of  the  pin  as  the  piston  rod 

moves  forward  or  backward.     On  account  of  these  alternating 

stresses  the  allowable  working  unit-stresses  should  be  taken  low 

in  designing  the  pin  and  crank  arm. 

The  shaft,  crank  arm,  and  pin  are  usually  forged,  the  whole 
being  one  piece  of  metal,  but  for  very  large  shafts  the  crank 
arm  may  be  bored  to  a  diameter  slightly  less  than  that  of  the 
end  of  the  shaft  and  then  shrunk  upon  it  (Art.  32).  Sometimes 
a  shaft  has  a  crank  arm  at  each  end,  it  being  then  called  a  double- 
throw  crank  shaft;  in  this  case  the  two  crank  arms  are  set  at 
right  angles  to  each  other,  in  order  that  the  action  of  the  two 
pressures  W  may  produce  a  more  uniform  twisting  moment  on 
the  shaft.  The  largest  crank  shaft  in  use  was  made  by  the 
Bethlehem  Steel  Company  in  1905;  it  is  37  inches  in  outside  and 


\W 


242  TORSION  OF  SHAFTS  CHAP,  x 

30  inches  in  inside  diameter,  the  crank  webs  are  64  inches  long, 
49  inches  wide  and  16  inches  thick,  and  the  solid  pins  are  19  inches 
in  diameter.  This  double-throw  crank  shaft  is  27  feet  long,  the 
two  crank  pins,  webs,  and  shaft  being  one  piece  of  steel  which 
was  forged  from  an  ingot  weighing  308  ooo  pounds,  and  the 
weight  of  the  finished  shaft  is  86  600  pounds. 

Prob.  97a.  The  crank  pin  CD  in  Fig.  97  is  8  inches  long  and  4  inches 
in  diameter,  the  pressure  W  being  60  ooo  pounds.  Compute  the  bear- 
ing unit-stress,  the  shearing  unit-stress,  and  the  flexural  unit-stress. 

Prob.  976.  Let  BC  be  14  inches  long,  4  inches  thick,  and  10  inches 
wide  at  B.  Compute  the  flexural  unit-stress  at  B  when  the  pressure 
of  60  ooo  pounds  acts  normal  to  BC. 

ART.  98.     A  TRIPLE-CRANK  SHAFT 

Double  and  triple  cranks  are  used  when  several  engines  are 
to  be  attached  to  the  same  shaft,  as  is  usual  in  ocean  steamers. 
With  the  triple  arrangement  the  cranks  are  set  at  angles  of  120 
degrees  with  each  other,  thus  securing  a  uniform  action  upon 
the  shaft.  Fig.  98  shows  one  of  these  cranks,  AB  and  EF  being 
portions  of  the  shaft  resting  in  journal  bearings,  CD  one  of  the 
crank  pins  to  which  the  connecting  rod  is  attached,  while  BC  and 
DE  are  the  crank  arms  or  webs  which  are  usually  shrunk  upon 
the  shaft  and  pins. 

The  complete  investigation  of  the  maximum  stresses  in  such 
a  crank  shaft  and  pin  is  one  of  much  difficulty.  A  brief  ab- 
stract of  such  an  investigation  will,  however,  here  be  given  for 
the  crank  pin.  There  are  three  cranks,  and  the  one  to  be  con- 
sidered is  the  nearest  to  the  propeller,  so  that  the  torsion  from 
the  other  two  cranks  is  transmitted  through  the  pin  CD.  This 
steel  crank  pin  is  hollow,  18  inches  in  outer  diameter  and  6  inches 
in  inner  diameter,  its  length  between  webs  being  2^  inches,  the 
thickness  of  each  web  12  inches,  and  the  distance  from  the  axis 
of  the  shaft  to  the  center  of  the  pin  being  30  inches.  The  three 
engines  transmit  7  200  horse-power  to  the  shaft  EF,  of  which 
4  800  horse-power  is  transmitted  through  the  shaft.  AB  and 
tnrough  the  crank  pin  CD.  The  maximum  pressure  W  brought 


ART.  98 


A  TRIPLE-CRANK  SHAFT 


243 


by  the  connecting  rod  upon  the  crank  pin  is  156  ooo  pounds.  It 
is  required  to  determine  the  stresses  when  the  crank  makes  80 
revolutions  per  minute. 

The  pressure  W  is  distributed  over  about  17  inches  of  the 
length  of  the  pin,  so  that  the  bearing  compressive  stress  on  the 
diametral  area  is  Si  =156  ooo/ 1 7*^12*^=765  pounds  per  square 
inch,  which  is  a  low  and  safe  value.  The  shearing  unit-stress 
due  to  Ty,  which  is  taken  as  uniformly  distributed  over  the  sec- 
tion  area  of  the  pin,  is  A  fa*~*~/&±^»  +f  j*^  (g) 

52=78  ooo/%n{iS>2  —  62)  =345  pounds  per  square  inch 
which  is  also  a  low  working  value  for  steel. 


Fig.  98 

The  shearing  stress  due  to  the  horse-power  transmitted 
through  BC  has  its  greatest  value  on  the  side  of  the  pin  farthest 
from  the  axis.  The  twisting  moment  Pp  due  to  this  4  800  horse- 
power is  found  from  formula  (91)  to  be, 

Pp  =  6T,  000X4  800/80  =  3  782  ooo  pound-inches 
and   this   is    equal  to   the  resisting  moment  of  the  crank  pin, 
or   to   S/i/Ci,   in    which   Ji   is   the    polar    moment    of   inertia 
of  the  cross-section  with  respect  to  the  axis  of  the  shaft  and  c\ 
is  the  distance  from  that  axis  to  the  side  of  the  pin  where  the 
stress  S  is  to  be  found.     Now,^  =  30  +  9  =  39  inches,  and  then  Ji  = 
0.0982 (i84-64)  +0.7854(182-62) X392.    Hence,  from  formula 
(90)  there  is  found, 

$3 =3  782  00^X39/354  000=420  pounds  per  square  inch 
which  is  the  shearing  stress  due  to  the  transmitted  power. 

The  flexure  of  the  pin  due  to  the  torsion  carried  through  it 
falls  under  the  case  discussed  in  Art.  65.  The  twisting  moment 
Pp  is  equivalent  to  a  force  P  acting  at  a  distance  of  30  inches 


244  TORSION  OF  SHAFTS  CHAP.  X 

from  the  shaft  and  normal  to  the  crank  arms;  the  value  of  P  is 
3782000/30  =  126100  pounds,  and  this  produces  a  bending 
moment  in  the  pin  which  may  be  taken  as  a  beam  fixed  at  both 
ends,  while  P  acts  in  opposite  directions  at  those  ends,  as  in  Fig. 
65a.  Hence  there  is  a  bending  moment  Mf  at  each  end,  oppo- 
site in  sign  but  equal  in  value,  and  the  moment  at  any  section  is 
M  =M'+Px;  but  when  oc=l  the  value  of  M  is  —  M.'  and  there- 
fore M '  =  ±  JP/,  which  is  the  maximum  bending  moment.  The 
flexure  formula  (41)  then  gives 

S±  =  M' .  c/I=6$  000X24X97 g-Vr(i84-64)  =  2  060 
which  is  the  flexural  stress  in  pounds  per  square  inch  due  to  the 
transmission  of  power  through  the  pin. 

All  of  these  stresses  are  light,  but  the  pin  is  necessarily  made 
heavier  than  they  would  require,  on  account  of  the  additional 
stresses  due  to  the  shrinking  of  the  web  upon  the  pin.  The  data 
here  given  are  not  sufficient  to  determine  these  shrinkage  stresses, 
but  the  discussion  in  Art.  154  indicates  that  there  is  a  radial  com- 
pressive  unit-stress  S  5  brought  by  the  web  upon  the  pin  of  proba- 
bly 3  ooo  pounds  per  square  inch,  and  that  this  is  accompanied 
by  a  tangential  compressive  unit-stress  SQ  of  about  4  ooo  pounds 
per  square  inch.  These  shrinkage  stresses  occur  also  in  the  fillet 
of  the  pin  on  the  inside  of  the  web  at  D,  where  all  the  other 
stresses  except  Si  concentrate.  In  Art.  179  it  will  be  shown  how 
these  several  values  may  be  combined  in  order  to  obtain  the  final 
maximum  tensile,  compressive,  and  shearing  stresses. 

Since  all  these  stresses  vary  in  direction  and  intensity  as  the 
shaft  revolves,  their  effect  is  more  injurious  than  steady  stresses, 
and  accordingly  the  factors  of  safety  should  be  high,  or  the  work- 
ing unit-stresses  low,  in  order  that  the  life  of  the  crank  pin  may 
not  be  short  (Art.  137).  Shocks  are  also  brought  upon  the  shaft 
of  an  ocean  steamei  when  the  propeller  at  the  stern  rises  out  of 
the  water  and  falls  back  again,  as  it  does  when  great  waves  cause 
longitudinal  oscillations,  and  these  shocks  also  require  that  the 
working  unit-stresses  shall  be  low. 

Prob.  98.  The  shaft  EF  in  Fig.  98  is  hollow,  the  inside  diameter 
being  8  inches  and  the  outside  diameter  24  inches.  Compute  the  flexu- 


ART.  99  NON-ClRCULAR   SECTIONS  245 

ral  unit-stress  S  when  7  200  horse-powers  are  transmitted  through  it 
while  it  is  making  65  revolutions  per  minute. 

i 
ART.  99.     NON-CIRCULAR  SECTIONS 

When  a  rectangular  bar  is  subject  to  torsion  the  general 
phenomena  are  the  same  as  those  described  in  Art.  89,  but  the 
distribution  of  the  shearing  stresses  over  the  section  area  is  ob- 
served to  be  different  from  that  in  a  circular  section.  In  Fig. 
169c  is  seen  a  steel  rectangular  bar,  ^Xi|X8J  inches,  which  has 
been  twisted  in  a  torsion  machine  through  an  angle  of  142  degrees. 
Before  the  test  the  surfaces  of  the  bar  were  blackened  and  two 
series  of  white  lines  drawn  thereon  at  right  angles  to  each  other; 
the  figure  plainly  shows  the  distortion  of  the  rectangles  into 
rhombuses,  the  greatest  distortion  in  the  right  angles  being 
at  the  middle  of  the  wide  side,  while  the  white  lines  remain  closely 
perpendicular  to  the  corner  lines  of  the  bar.  From  many  ex- 
periments of  this  kind  it  is  concluded,  since  stresses  are  propor- 
tional to  distortions,  that  the  shearing  unit-stress  at  the  middle 
of  the  wide  side  is  greater  than  that  at  the  middle  of  the  narrow 
side  of  the  rectangular  bar,  and  that  there  are  no  shearing  stresses 
along  the  corners. 

When  lines  are  ruled  upon  the  surface  of  a  round  bar  it  is 
observed  that  the  distortion  of  the  angles  is  the  same  on  all  parts 
of  the  surface,  and  hence  the  shearing  unit-stress  is  uniform 
over  that  surface;  this  is  the  basis  of  the  torsion  formula  deduced 
in  Art.  90.  For  a  bar  of  elliptical  section,  however,  it  is  observed 
that  the  distortion  of  the  angles  is  greatest  on  the  flatter  side 
of  the  section;  hence  the  shearing  unit-stress  is  there  the  greatest, 
and  the  torsion  formula  (90)  does  not  apply  to  the  elliptical  bar; 
in  fact,  that  formula  applies  only  to  circular  sections,  and  it 
should  not  be  used  for  other  sections  except  for  approximate 
investigations.  The  correct  torsion  formulas  for  elliptical  and 
rectangular  sections  will  now  be  deduced. 

Let  Fig.  990  represent  a  section  of  an  elliptical  bar  which 
is  subject  to  a  twisting  moment  Pp  from  a  force  P  acting  in  a 
plane  normal  to  the  axis  and  at  a  distance  p  from  that  axis.  Let 


246  TORSION  OF  SHAFTS  CHAP,  x 

m  be  the  major  axis  and  n  the  minor  axis  of  the  ellipse,  let  y 
and  x  be  the  vertical  and  horizontal  coordinates  of  any  point 
on  the  circumference  of  the  ellipse  with  respect  to  m  and  n  as 
coordinates  axes.  Let  Si  and  S2  be  the  shearing  unit-stresses  at 
the  extremities  of  the  minor  and  major  axes,  and  S  the  shearing 
unit-stress  at  the  point  where  coordinates  are  y  and  x]  these 
stresses  are  tangential  to  the  circumference.  Let  Sr  and  S"  be 
the  components  of  S  parallel  to  Si  and  ^2,  and  let  %  be  the  angle 
which  S  makes  with  Si]  then  Sf  =  S  cosj  and  S"  =S  sin^,  whence 
S"/S' =tan.%.  The  equation  of  the  ellipse  is  m2y2  +  n2x2  =  {m2n2, 
and  by  differentiation  there  is  found  dy/dx=  —  n2x/m2y  =tsin%] 
accordingly  S"/S'  =  n2x/m2y  and  it  thus  appears  that  the  com- 
ponents S"  and  S'  are  proportional  to  their  distances  from  the 
coordinate  axes  n  and  m.  When  the  elastic  limit  of  the  material 
is  not  exceeded,  the  same  relation  must  hold  between  the  com- 
ponents of  the  unit-stress  at  any  point  within  the  ellipse,  for 
here,  as  in  the  circle,  the  unit-stresses  along  any  radius  vector 
vary  proportionally  as  their  distances  from  the  center. 


Fig.  990  Fig.  996 

Now  let  yi  and  xi  refer  to  any  point  within  the  ellipse  and 
let  m\  and  n\  be  the  axes  of  an  ellipse  passing  through  that  point3 
the  ratio  m\/n\  being  equal  to  m/n.  Let  Sx  and  Sy  be  the  com- 
ponents of  the  unit-stress  at  that  point,  Sx  being  parallel  to  Si 
and  Sy  to  S2j  then  also  Sy/Sx=n^x\/m-fy\.  Let  da  be  the 
element  of  area  at  the  given  point;  then  Sxyida-kSyXida  is  the 
resisting  twisting  moment  of  the  stress  on  that  area,  and  the  sum 
of  all  similar  expressions  for  all  elements  of  the  area  equals  the 
twisting  moment,  or 

2Sxyi  da  +  ISyXi  da  =  Pp 

Substituting  in  this  for  Sy  its  value  Sx  .  f»i2#i/*»iayi  and  then  for 
Sx  its  value  S'.  y\/y,  and  also  for  m\/n\  its  value  m/n,  the  com- 


ART.  9&  NON-ClRCULAR    SECTIONS  247 

ponent  S'  is  founa.  Similarly  substituting  for  Sx  its  value 
Sy  .  mi2yi/ni2Xi  and  then  for  Sy  its  value  S"  .  x\/x,  the  com- 
ponent S"  is  found.  In  both  cases  2y\2da  is  the  moment  of  in- 
ertia /i  of  the  area  of  the  ellipse  with  respect  to  the  major  axis  m, 
while  Zx\2da  is  the  moment  of  inertia  /2  with  respect  to  the 
minor  axis  n.  Thus  are  found, 

Pp.m2y  Pp.n2* 


which  are  the  components  of  the  shearing  unit-stress  S  at  any 
point  on  the  circumference  whose  coordinates  are  y  and  x.  When 
y  =o  and  x  =  Jw,  the  horizontal  component  S'  is  o  and  the  vertical 
component  S"  becomes  S2  in  the  figure  ;  when  x  =  o  and  y  =  \n,  the 
vertical  component  S"  is  zero  and  the  horizontal  component  be- 
comes Si  in  the  figure.  The  ratio  Si/S%  is  hence  equal  to  m2n/mn2 
or  to  m/n,  that  is,  the  shearing  unit-stresses  at  the  ends  of  the  axes 
are  inversely  proportional  to  the  lengths  of  those  axes.  The 
greatest  unit-stress  hence  occurs  at  the  ends  of  the  minor  axis, 
while  the  least  unit-stress  on  the  circumference  occurs  at  the  ends 
of  the  major  axis;  hence  Si  is  the  unit-stress  to  be  used.  The 
rectangular  moment  of  inertia  of  the  ellipse  with  respect  to  its 
major  axis  is  /i  =  -fax-inn3  and  that  with  respect  to  the  minor 
axis  is  I2  =  fanm3n  (Art.  43).  Substituting  these  values  in  the 
above  value  of  Sr  and  making  y  =  ^n,  there  results, 

or  Pp  =xmn2&Si 

which  are  the  formulas  for  discussing  a  bar  of  elliptical  cross- 
section  under  torsion.  When  m=n  =  d,  the  ellipse  becomes  a 
circle,  and  the  formula  reduces  to  that  deduced  for  the  circle 
in  Art.  92. 

For  the  rectangle  in  Fig.  996,  the  exact  discussion  is  much 
more  difficult  because  the  stresses  are  not  proportional  to  their 
distances  from  the  axis  except  along  the  median  lines.  On  any 
side  the  stress  is  greatest  at  the  middle  and  varies  approximately 
as  the  ordinates  of  a  parabola  toward  each  corner  where  it  be- 
comes zero;  the  stress  at  the  middle  of  the  broad  side  is  greater 
than  that  at  the  middle  of  the  narrow  side,  as  in  the  elliptical 
section.  Let  m  and  n  be  the  long  and  short  sides  of  the  rectangle, 


248  TORSION  OF  SHAFTS  CHAP.  X 

and  let  x  and  y  be  coordinates  of  any  point  within  the  rect- 
angle with  respect  to  axes  through  the  center,  x  being  parallel 
to  m  and  y  to  n.  Then  the  above  facts  of  experiment  indicate 
that  the  variation  of  stresses  throughout  the  rectangle  is  closely 
expressed  by 


where  Si  and  S2  are  the  unit-stresses  at  the  middle  of  the  long 
and  short  sides,  respectively,  and  Sx  and  Sy  are  the  components 
parallel  to  Si  and  S2  of  the  unit-stress  at  any  point.  The  funda- 
mental equation  between  resisting  and  twisting  moment  is  also 
the  same  as  before;  replacing  the  sign  of  summation  by  that  of 
integration  and  da  by  dx  dy,  it  becomes, 

fsxy  dx  dy+fSyX  dx  dy=Pp 

Inserting  in  this  the  values  of  S  x  and  Sv  and  integrating,  first 
between  the  limits  -\-\n  and  —  \n  for  y,  and  second  between  the 
limits  +%m  and  —  \m  for  x,  it  becomes  %mn(Sin  +  S2w)  =  Pp. 
Now,  as  in  the  ellipse,  81/82  equals  m/n,  and  accordingly  S^m 
equals  S\n\  whence,  since  S\  is  the  greatest  unit-stress, 

Si  =  lPp/mn2  or  Pp=%mn2Si 

are  the  formulas  for  discussing  rectangular  bars  under  torsion. 
Apply  to  a  square  bar  of  side  d  by  making  mn2  equal  to  d3. 

A  comparison  of  the  strength  of  solid  round  and  square  shafts 
is  now  readily  made  from  the  values  of  the  twisting  moments 
derived  in  Art.  92  and  in  the  preceding  paragraph: 

for  a  round  shaft  ,          Pp  =  -fairSd?  =0.1  964$$* 
for  a  square  shaft,         Pp—  %Sd3=o.2222Sd3 

and  accordingly  the  strength  of  a  square  shaft  of  side  d  is  13 
percent  greater  than  that  of  a  round  shaft  of  diameter  d,  the  shear- 
ing unit-stress  being  the  same  in  the  two  cases.  When  power  is 
transmitted  by  a  square  shaft,  Pp  is  to  be  replaced  by  63  o^oH/n 
for  English  measures  and  then  are  found 

S  =284000^/^3  or  d=6$.>j(H/nS)* 

which  are  the  formulas  for  the  investigation  and  design  of  solid 
square  shafts  similar  to  those  of  (92)  for  solid  round  shafts. 


ART.  99  NON-ClRCULAR  SECTIONS  249 

The  angle  of  twist  produced  in  a  non-circular  section  by  a 
given  twisting  moment  cannot  be  deduced  by  the  method  given 
in  Art.  93  for  round  shafts,  on  account  of  the  irregular  distribution 
of  stresses.  The  investigations  of  Saint  Venant,  who  was  the  first 
to  establish  the  correct  theory  of  torsion,  give  the  following  value 
.for  the  angle  of  twist  for  a  square  shaft,  that  for  the  round  shaft 
being  derived  from  Art.  93  : 

for  a  round  shaft,          <£  =  lo.iSPp  .  l/Fd* 
for  a  square  shaft,         </>=  j.uPp  .  l/Fd4 

and  hence  the  angle  of  twist  of  a  round  shaft  is  43  percent  greater 
than  for  a  square  one.  These  values  of  </>  are  in  radians;  when 
the  angle  is  desired  in  degrees,  it  should  be  remembered  that  one 
radian  is  equivalent  to  57.3  degrees. 

All  the  formulas  of  this  article  are  valid  only  when  the  great- 
est Shearing  unit-stress  does  not  exceed  the  elastic  limit  of  the 
material.  The  formula  S  =  %Pp/mn2  may,  however,  be  used  to 
compute  the  so-called  torsional  modulus  of  rupture  or  torsional 
strength  when  a  rectangular  bar  is  ruptured  by  torsion,  n  being 
the  short  side  of  the  rectangle.  When  rupture  occurs  in  the 
twisting  of  such  a  bar,  it  usually  begins  at  the  middle  of  the  flat 
side,  so  that  even  in  this  extreme  case  there  is  little  shearing  stress 
on  the  corners.  Cracks  occurring  on  the  corners  are  to  be  attrib- 
uted to  tensile  stresses  which  accompany  the  elongation  due  to 
the  change  of  a  straight  line  into  a  helix.  For  example,  the 
square  steel  bar  in  Fig.  188  was  11}  inches  long  between  the  jaws 
of  the  torsion  machine,  and  this  length  was  not  increased  by  the 
twist  of  900  degrees.  The  side  of  the  square  being  0.75  inches, 
the  length  of  its  diagonal  is  1.06  inches,  the  length  of  a  circum- 
ference of  900  degrees  described  by  the  corner  is  8.33  inches, 
and  the  length  of  the  helix  is  14.40  inches;  hence  the  increase  in 
length  of  the  corner  line  was  2.65  inches  and  the  percentage  of 
elongation  was  nearly  23  percent,  which  is  less  than  the  ultimate 
elongation.  The  specimen  broke  by  shearing  at  one  end  under 
a  twisting  moment  of  5  850  inch-pounds,  so  that  the  computed 
torsional  strength  of  the  steel  is  5  =9X5  850/2X0.753  =62  400 
pounds  per  square  inch. 


260  TORSION  OF  SHAFTS  CHAP,  x 

Prob.  99a.  Check  the  computations  in  the  last  paragraph. 

Prob.  996.  Compare  the  strength  of  a  round  shaft  with  that  of  a 
square  one  having  the  same  area  of  cross-section. 

Prob.  99c.  A  square  wooden  .shaft  for  a  water  wheel  is  12  inches 
square  and  transmits  36  horse-powers  at  9  revolutions  per  minute. 
Compute  its  factor  of  safety. 

Prob.  99 d.  The  rectangular  bar  of  medium  steel  in  Fig.  169c  was 
twisted  through  an  angle  of  28.5  degrees  by  a  twisting  moment  of 
2  800  inch-pounds,  the  length  of  the  bar  being  8J  inches,  its  thickness 
J  inch,  and  its  width  ij  inches.  Compute  the  shearing  unit-stress  for 
these  data,  and  determine  whether  or  not  the  shearing  elastic  limit  of 
the  material  was  exceeded. 


ART.  100  STRESSES  DUE  TO  TEMPERATURE  251 


CHAPTER  XI 

APPARENT  COMBINED  STRESSES 
ART.  100.     STRESSES  DUE  TO  TEMPERATURE 

Several  axial  loads  produce  the  same  unit-stress  in  a  bar  as 
a  single  load  equal  to  their  algebraic  sum  and  the  change  of 
length  which  is  that  due  to  this  load.  When  the  bar  is  thus 
stressed  at  a  certain  temperature,  a  change  in  temperature  usually 
causes  the  existing  stress  to  become  greater  or  less.  When  a  bar 
is  free  to  expand  or  contract  under  a  rise  or  fall  of  temperature, 
there  occurs  a  change  of  length  which  is  unaccompanied  by 
internal  stress,  for  in  this  case  there  is  no  external  force  and  stress 
is  an  internal  resistance  to  an  applied  external  force.  If  this 
change  of  length  is,  however,  prevented  from  occurring  by  fast- 
ening the  ends  of  the  bar,  there  is  produced  an  internal  stress 
which  is  the  same  as  that  which  would  be  caused  by  an  external 
force  which  would  shorten  or  lengthen  the  free  bar  the  same 
amount  that  it  has  expanded  or  contracted.  For  example,  let  a 
free  steel  bar  100  inches  long  become  99.9  inches  long  under  a 
certain  fall  of  temperature,  then  no  internal  stress  is  caused  by 
this  change  in  length ;  to  bring  this  bar  back  to  its  original  length, 
there  must  be  effected  the  unit-elongation  0.1/100=0.001  and 
Art.  10  shows  that  this  will  require  a  tension  of  o.ooi  X  30  ooo  ooo  = 
30  ooo  pounds  per  square  inch.  Hence,  if  this  bar  is  prevented 
from  shortening  under  the  given  fall  of  temperature,  a  tensile 
unit-stress  of  30  ooo  pounds  per  square  inch  is  produced  in  every 
cross-section. 

Let  e  be  the  change  per  unit  of  length  which  occurs  when  a 
bar  is  free  to  expand  or  contract,  y  the  coefficient  of  linear  ex- 
pansion or  change  per  unit  of  length  for  a  rise  or  fall  of  one  degree 
Fahrenheit,  and  /  the  number  of  degrees  of  rise  or  fall;  also  let 
S  be  the  unit-stress  which  will  occur  if  the  bar  is  prevented  from 
expanding  or  contracting,  and  E  be  the  modulus  of  elasticity  of 


252  APPARENT  COMBINED  STRESSES  CHAP.  XI 

the  material.  Then  from  the  preceding  paragraph  and  from 
Art.  9, 

£  =  yt  S=eE  S=ytE  (100) 

Et  is  thus  seen  that  the  unit-stress  due  to  change  of  temperature 
is  independent  of  the  length  of  the  bar;  if  a  is  the  section  area 
of  the  bar,  the  total  stress  in  each  section  due  to  the  change  in 
temperature  is  aS. 

The  following  are  average  values  of  the  coefficients  of  linear 
expansion  for  one  degree  of  the  Fahrenheit  scale : 

for  brick  and  stone,  7]  =0.000  0050 

for  cast  iron,  77=0.000  0062 

for  wrought  iron,  77=0.000  0067 

for  steel,  77=0.0000065 

From  these  coefficients  the  change  per  unit  of  length  due  to  a 
rise  or  fall  of  /  degrees  is  readily  computed  or  the  unit-stress  5 
may  be  directly  found.  This  temperature  stress  is  to  be  added 
to  or  subtracted  from  the  tensile  or  compressive  stress  due  to  the 
applied  forces  on  the  bar. 

As  an  example  consider  a  wrought-iron  tie  rod  20  feet  in 
length  and  2  inches  in  diameter  which  is  screwed  up  to  a  ten- 
sion of  9  ooo  pounds  in  order  to  tie  together  two  walls  of  a  build- 
ing. Let  it  be  required  to  find  the  stress  in  the  rod  when  the 
temperature  falls  10  degrees  Fahrenheit.  Here, 

5=0.0000067X10X25000000  =  1  675  pounds 

and  the  stress  due  to  change  of  temperature  is  3.14X1675  = 
5  200  pounds,  so  that  the  total  tensile  stress  in  the  bar  becomes 
9  coo +  5  200  =  14  200  pounds.  For  a  rise  of  10  degrees  Fahren- 
heit, the  tensile  stress  in  the  bar  becomes  9000-5200=3800 
pounds. 

It  is  seen  from  the  above  that  the  unit-stress  caused  in  a  steel 
bar  by  a  change  of  one  degree  Fahrenheit  is  about  200  pounds 
per  square  inch,  so  that  a  change  of  100  degrees  might  cause  a 
stress  of  20  ooo  pounds  per  square  inch  if  no  provision  were 
made  for  allowing  the  bar  to  change  its  length.  Steel  bridges 
usually  rest  on  rollers  at  one  end  so  that  change  in  length  may 
occur  under  change  of  temperature  and  thus  stresses  due  to 


ART.  101  BEAMS  UNDER  AXIAL  FORCES  253 

temperature  be  prevented.  When  railroad  rails  are  laid  in  cold 
weather,  it  is  customary  to  leave  the  ends  about  J  inch  apart,  so 
that  there  may  be  room  for  expansion  when  the  warm  weather 
comes;  the  holes  in  webs  of  the  rails,  through  which  bolts  pass 
to  connect  the  splice  bars,  are  made  oval  instead  of  round  so  that 
the  rails  may  be  free  to  expand  and  contract. 

Prob.  100.  What  is  the  change  in  length  of  a  steel  railroad  rail  60 
feet  long  when  the  temperature  rises  from  — 10  to  +80  degrees  Fah- 
renheit? If  the  rail  weighs  95  pounds  per  yard,  what  force  is  required 
to  prevent  this  expansion,  and  what  compressive  unit-stress  will  it 
cause  in  the  rail? 

ART.  101.    BEAMS  UNDER  AXIAL  FORCES 

A  normal  stress  is  one  acting  normally  to  the  section  area 
of  a  bar,  and  this  must  be  either  tensile  or  compressive  (Art.l). 
When  several  applied  axial  forces  act  upon  a  bar  each  produces 
a  stress  on  the  section  area  and  the  sum  of  these  stresses  must 
equal  the  total  load.  Hence  the  combination  of  normal  stresses 
is  made  by  simple  addition,  if  all  are  tensile  or  all  are  compressive; 
when  some  are  tensile  and  others  compressive,  their  algebraic 
sum  is  to  be  taken. 

A  beam  under  transverse  loads  has  normal  stresses  of  tension 
on  one  side  of  its  neutral  surface  and  normal  stresses  of  com- 
pression on  the  other  side  (Art.  39).  When  the  beam  is  under 
an  axial  tension  P  which  is  uniformly  distributed  over  the  section 
area  a,  the  unit-stress  P/a  is  to  be  added  to  each  of  the  flexural 
tensile  unit-stresses  and  be  subtracted  from  each  of  the  flexural 
compressive  unit-stresses.  Thus,  if  the  unit-stresses  due  to  the 
flexure  on  the  tensile  and  compressive  sides  of  the  beam  are 
Si  and  $2,  then  Si  +  P/a  is  the  tensile  unit-stress  due  to  flexure 
and  longitudinal  tension,  while  S^  —  P/a  is  the  compressive 
unit-stress  due  to  flexure  and  axial  tension. 

An  approximate  method  of  finding  S\  and  £2  is  by  mean 
of  the  flexure  formula  (41),  which  is  applied  to  the  transverse 
loads  just  as  if  the  axial  tension  were  not  acting.  For  example, 
let  it  be  required  to  find  the  factor  of  safety  of  a  1 2-inch  I  beam 


254  APPARENT  COMBINED  STRESSES  CHAP.  XI 

of  6  feet  span,  weighing  55  pounds  per  linear  foot,  which  carries 
a  uniform  load  of  i  200  pounds  besides  its  own  weight,  when  sub- 
ject to  an  axial  tension  of  80  ooo  pounds.  The  flexure  formula  is 
Si  .  I/c=M;  from  Table  6  the  section  factor  I/c  is  53.5  inches3; 
from  Art.  38  the  bending  moment  M  is  1530X6/8  =  1147.5 
pound-feet;  hence  the  tensile  unit-stress  on  the  lower  side  of 
the  simple  beam  is  Si  =  25 7  pounds  per  square  inch.  Table  6 
gives  the  section  area  0  =  16.18  square  inches,  and  hence  the 
unit-stress  due  to  the  axial  tension  is  P/a  =  80  000/16.18  =4  940 
pounds  per  square  inch.  Hence  on  the  lower  side  of  the  beam, 
the  total  tensile  stress  is  260  +  4940  =  5200  pounds  per  square 
inch,  and  the  factor  of  safety  is  60  000/5  2O°  =  I  zi-  OR  tne  upper 
side  of  the  beam,  the  stress  is  tensile,  since  P/a  is  greater  than  S23 
and  its  value  is  4940  —  260=4680  pounds  per  square  inch. 

When  the  axial  force  on  the  beam  is  compression,  a  similar 
approximate  method  may  be  followed,  the  compressive  unit- 
stress  Si  on  the  concave  side  being  found  from  the  flexure  formula, 
while  the  unit-stress  due  to  the  load  is  found  from  P/a,  if  the 
beam  is  short,  or  from  the  column  formula  (80)  if  its  length 
exceeds  ten  or  twelve  times  Us  least  thickness. 

A  rafter  of  a  roof  is  a  case  of  combined  compression  and 
flexure,  for  a  rafter  is  under  compression  from  the  forces  that  act 
upon  its  ends  and  under  flexure  from  its  weight  and  that  of  the 
roof  covering.  In  many  cases  the  approximate  method  here 
outlined  is  sufficient  for  its  investigation.  Let  the  section  oi 

the  rafter  be  rectangular,  b  being 
its  width,  d  its  depth,  /  the  length, 
w  the  uniform  load  per  linear 
unit,  and  <f>  the  angle  of  inclina^ 
Fig.  101.  tion.  To  find  the  horizontal  re- 

action H,  the  center  of  moments  is  taken  at  the  lower  end,  and, 

H  .  I  sin</>  =  wl .  i/  cose/)  whence    H  =  \wl  cot</> 

For  any  section  area  at  the  distance  x  from  the  upper  end  of  the 
rafter,  the  flexure  formula  gives  the  flexural  unit-stress, 

6M    6(Hx  sin(f> 
^l=~bd?~  bd? 


ART.  102  FLEXURE  AND  COMPRESSION  255 

and  the  uniform  compressive  unit-stress  is, 
P    H  cosd>+wx  sin0 
a  bd 

The  total  compressive  unit-stress  on  the  upper  fiber  hence  is, 


bd2  2bd  bd 

By  the  usual  method  this  is  found  to  be  a  maximum  when 


and  substituting  this,  the  maximum  unit-stress  is, 
_  3  wl2  cos(f>     wl  cosec0 


2bd 

which  formula  may  be  used  to  investigate  or  to  design  common 
^afters  subject  to  uniform  loads. 

In  any  inclined  rafter,  let  P  denote  all  the  load  above  a  sec- 
tion distant  x  from  the  upper  end.  Then  reasoning  as  before 
the  greatest  unit-stress  for  that  section  is  found  to  be, 

Mc    P  sm(>     H 


x 

la  a 

from  which  Sx  may  be  computed  for  any  given  case. 

Prob.  101  a.  Find  the  size  of  a  square  wooden  simple  beam  of  12 
feet  span  to  carry  a  load  of  300  pounds  at  the  middle  when  it  is  also 
subject  to  a  longitudinal  tension  of  2  ooo  pounds,  the  allowable  tensile 
stress  being  i  ooo  pounds  per  square  inch. 

Prob.  10  Ib.  A  roof  with  two  equal  rafters  is  40  feet  in  span  and 
15  feet  in  height.  The  wooden  rafters  are  4  inches  wide  and  each 
carries  a  load  ot  450  pounds  at  the  middle.  Find  the  depth  of  the 
rafter  so  that  S  may  be  700  pounds  per  square  inch. 

ART.  102.     FLEXURE  AND  COMPRESSION 

Let  a  beam  be  subject  to  flexure  by  transverse  loads  and 
also  to  an  axial  compression  in  the  direction  of  its  length.  If  the 
longitudinal  compression  is  not  large,  the  combined  maximum 
stress  due  to  flexure  and  compression  may  be  computed  by  the 
approximate  method  of  Art.  101.  It  is  clear,  however,  that 
if  the  compression  is  large  the  deflection  of  the  beam  will  be 


256  APPARENT  COMBINED  STRESSES  CHAP.  XI 

increased  by  it,  and  hence  the  effective  bending  moment  and 
maximum  fiber  stresses  will  be  greater  than  given  by  that  method. 
A  closer  approximation  will  now  be  established. 

Let  P  be  the  axial  compressive  force  and  M  the  bending 
moment  of  the  flexural  forces.  Let  M\  be  the  actual  bending 
moment  for  the  dangerous  section  where  the  actual  deflection  is 
/i  ;  this  is  greater  than  M,  on  account  of  the  moment  Pfi  of  the 
force  P,  or  Mi=M  +  Pf\.  Now  the  maximum  fiber  unit-stress 
Si  which  results  from  this  moment  M\  is,  from  (41), 

SI-MI.  c/i-*(M+pfdc/i 

where  /  is  the  moment  of  inertia  of  the  cross-section  and  c  the 
distance  from  the  neutral  axis  to  the  remotest  fiber  on  the  com- 
pressive side.  The  value  of  f\  may  be  expressed  in  terms  of  Si, 
regarding  f\  to  vary  with  Si  in  the  same  manner  as  for  a  beam 
subject  to  no  axial  compression.  Inserting  then  for  f\  its  value 
from  (56),  and  solving  for  Si,  gives, 


*>->?/(-£) 


where  a  and  ft  are  numbers  that  depend  upon  the  arrangement 
of  the  ends  and  the  kind  of  loading  of  the  beam;  for  a  simple 
beam  uniformly  loaded  the  value  of  ft/a  is  9.6;  for  a  simple 
beam  with  load  at  the  middle  ft  /a  is  12. 

The  maximum  compressive  unit-stress  on  the  concave  side 
of  the  beam  is  S=Si+P/a.  For  example,  let  a  simple  wooden 
beam  8  feet  long,  10  inches  wide,  and  9  inches  deep  be  under 
an  axial  compression  of  40  ooo  pounds,  while  at  the  same  time  it 
carries  a  total  uniform  load  of  4000  pounds.  Here  M  =  ^Wl 
=  48000  pound-inches,  c=4^  inches,  I  =-fabd?=6o'j%  inches4, 
/  =  96  inches,  P  =40  ooo  pounds,  a  =  8,  ft  —f1,  and  E  =  i  500  ooo 
pounds  per  square  inch.  Inserting  these  values  in  the  formula,  the 
value  of  Me  1  1  is  356,  and  then  the  final  flexural  stress  Si  is  found 
to  be  371  pounds  per  square  inch.  The  compressive  unit-stress 
due  directly  to  P  is  P/a  =  40  000/90  =  444,  so  that  the  total  stress 
S  =  37  1  +444  =  815  pounds  per  square  inch. 

Another  method,  which  has  a  more  satisfactory  theoretical 
basis,  is  to  consider  the  flexural  stress  due  to  P  as  resulting  from 


A.RT.  102  FLEXURE  AND  COMPRESSION  257 

its  eccentricity  with  respect  to  the  section  at  the  middle  of  the 
beam.  Under  the  action  of  its  own  weight  a  simple  beam  has 
the  deflection/;  this  is  increased  by  the  action  of  the  eccentric 
load,  and  from  Art.  87,  the  total  deflection  is  fsecO  where  6 
denotes  the  quantity  \(P12/ET)^  for  a  beam  with  supported 
ends.  As  before,  the  total  flexural  unit-stress  S\  is  given  by 
(M  +  Pfi)c/I}  in  which  /i  is  to  be  replaced  by  /  sec#  ;  also  replacing 
I  by  ar2,  where  r  is  the  radius  of  gyration  of  the  section  in  the 
plane  of  bending,  the  total  flexural  unit-stress  Si  at  the  middle 
of  the  concave  side  of  the  beam  is, 


(102)' 

in  which  6  has  the  values  given  in  Art.  87  for  different  arrange 
ments  of  the  ends  of  the  beam  ;   sec#  can  be  found  from  a  trigono- 
metric table  or  from  the  series  in  Art.  87. 


Fig.  102a  Fig.  102& 

To  illustrate  this  method,  let  the  data  of  the  above  numerical 
example  be  again  used.  The  value  of  Me /I  is  356  pounds  per 
square  inch  and  this  is  the  flexural  unit-stress  due  to  the  uniform 
load  alone.  The  last  term  of  the  second  member  gives  the 
flexural  unit-stress  due  to  the.  moment  of  P;  here  £=4.5 
inches,  from  Art.  55  the  deflection  due  to  W  is  f=$Wl3/3&4EI  = 
0.0506  inches,  r2  =607.5/90  =6.75  inches2,  0  =  \(PPj '£/)* =  0.3184 
radians  =  18°  15',  sec0  =  1.053,  and  lastly  (P/a)(cf/r2)  sec#  =  i6 
pounds  per  square  inch.  The  total  flexural  stress  S-\  is  then 
356  +  16  =  372  and  total  compressive  stress  S  is  372+444  =  816 
pounds  per  square  inch,  which  is  practically  the  same  as  that 
previously  found.  The  two  methods  give,  in  fact,  closely  the 
same  results  for  the  common  cases  which  occur  in  practice. 

While  the  above  methods  are  satisfactory  in  regard  to  numerical 
results,  a  more  exact  method  of  dealing  with  combined  flexure 
and  compression  is  by  help  of  the  elastic  curve.  For  the  common 
case  of  a  simple  beam  loaded  uniformly  with  w  per  linear  unit 


258  APPARENT  COMBINED  STRESSES  CHAP.  XI 

and  under  the  longitudinal  compression  P,  the  bending  moment 
for  any  section  distant  x  from  the  left  end  of  the  beam  is 
fywlx  —  %ivx2  +  Py,  and  the  differential  equation  of  the  elastic 
curve  of  the  beam  is, 


where  the  negative  sign  of  the  moment  is  used  because  of  the 
theoretic  requirement  that  y  and  the  second  derivative  must 
have  opposite  signs  when  the  curve  is  concave  to  the  axis  of  x. 
By  two  integrations  there  is  found, 

wlx     wx2      wEI  /cos  (2x—  1)6/1       \ 
~^P  +  ^F  +  ~P2~(        ^6~~        V 

in  which,  as  before,  6  is  an  abbreviation  for  J(P/2AE7)*.     In 

this  equation  of  the  elastic  curve,  let  x  =  J/,  then  y  =/i  and  then, 

jpa/j  =  _  %wl2P+wEI(sec6  -  1) 

which  gives  the  deflection  of  the  beam  due  to  both  the  uniform 
load  and  the  longitudinal  compression.  Inserting  the  value  of 
the  deflection  /i  in  the  expression  for  Si  at  the  beginning  of  this 
article,  there  is  found, 

Si  =  (wcE/P)  (secfl  -  1)  (102)" 

as  the  flexural  unit-stress  at  the  middle  of  the  concave  side. 

To  illustrate  this  method  let  the  data  of  the  above  numerical 
example  be  again  used.  Here  iv  =4  000/96  pounds  per  lineal- 
inch,  and  the  other  quantities  as  before,  also  seed  =  1.0530;  and 
then  (102)"  gives  Si  =373,  whence  5  =  373+444  =  817  pounds  per 
square  inch,  which  is  practically  the  same  as  found  by  the  other 
methods.  The  rough  method  of  Art.  101  gives  S  =800  pounds 
per  square  inch,  and  in  many  cases  this  method  may  be  used  to 
obtain  results  which  are  sufficiently  precise. 

When  w=o  in  formula  (102)",  the  case  is  that  of  a  column 
under  the  axial  load  PI  and  both  /i  and  Si  are  zero  when  P  is 
less  than  the  value  given  by  Euler's  formula  (Art.  78),  and  inde- 
terminate when  P  reaches  that  value.  On  the  other  hand,  when 
P=o,  the  case  is  that  of  a  simple  beam  uniformly  loaded,  and 
it  may  be  shown  that  the  above  formula  for  /i  will  reduce  to 
,  while  that  for  Si  will  reduce  to 


ART.  103  FLEXURE  AND  TENSION  259 

Prob.  1020.  Prove,  by  using  the  method  of  the  differential  calculus 
for  evaluating  indeterminate  quantities,  that  the  statement  in  the  last 
sentence  is  correct. 

Prob.  102ft.  A  wooden  cantilever  beam,  10  inches  wide  and  4  feet 
long,  carries  a  uniform  load  of  500  pounds  per  foot  and  is  subjected  to 
a  longitudinal  compression  of  40  ooo  pounds.  Find  the  depth  of  the 
beam  so  that  the  maximum  compressive  unit-stress  may  be  about  8oc 
pounds  per  square  inch. 


ART.  103.    FLEXURE  AND  TENSION 

Let  a  beam  be  subject  to  flexure  by  transverse  loads  and  then 
to  a  tension  in  the  direction  of  its  length.  The  effect  of  the  ten- 
sion is  to  decrease  the  deflection  from  /  to  /i,  and  thus  also  the 
tensile  flexural  stress.  If  M  is  the  bending  moment  of  the  trans- 
verse loads,  and  MI  that  of  the  combined  flexure  and  tension, 
then  Mi=M  —  Pf\.  Let  Si  be  the  resulting  flexural  unit-stress 
on  the  fiber  most  remote  from  the  neutral  surface  on  the  tensile 
side;  then  formula  (102)  gives  Si,  if  the  minus  sign  in  the  denomi- 
nator is  changed  to  plus.  Accordingly, 


and  Si+P/a  is  the  total  unit-stress  on  the  convex  side  of  the 
beam  resulting  from  the  combined  flexure  and  tension. 

As  an  example,  take  a  steel  eye-bar  18  feet  long,  i  inch  thick, 
and  8  inches  deep,  under  a  longitudinal  tension  of  80  ooo  pounds, 
E  being  29  ooo  ooo  pounds  per  square  inch.  The  weight  of  the 
bar  is  490  pounds,  and  M  =  JX49oXi8Xi2  =13  230  pound- 
inches.  Also  €=4  inches,  7  =  42.67  inches4,  ft  /a  =9.6,  P  = 
80  ooo  pounds,  /  =  2i6  inches.  Then  the  value  of  Me/  1  is  i  240, 
and  the  flexural  tensile  stress  Si  is  943  pounds  per  square  inch. 
Finally,  the  total  tensile  stress  on  the  convex  side  at  the  middle 
of  the  beam  is  5  =  943  +  10000  =  10943  pounds  per  square  inch. 

Formula  (102)'  also  applies  to  combined  flexure  and  tension 
by  changing  the  sign  of  P  from  plus  to  minus.  Here  6  becomes 
imaginary  and  the  circular  secant  becomes  the  hyperbolic  secan* 


260  APPARENT  COMBINED  STRESSES  CHAP.  XI 

which  is  designated  by  sech  6.    Then, 

Si-^-f.^secW  (103)' 

in  which  sech#  may  be  computed  from  2/(e°  +  e~e),  where  e  is 
the  base  of  the  Naperian  system  of  logarithms  and  6  denotes  the 
real  positive  number  J(P/2/E/)*.  For  instance,  let  the  data  of 
the  last  paragraph  be  again  considered.  The  value  of  Me  /I  is 
i  240  pounds  per  square  inch,  which  is  the  flexural  unit-stress 
due  to  uniform  load  alone.  Also  0=4  inches,  r2  =42. 67/8  = 
5.334  inches,  /=  $WP/$4EI  =0.05 2  inches,  0  =  ±(P12/EI)*  = 
0.868,  ^  =  2.7i8°-868  =  2.383,  e~e  =  1/^=0.420,  sech0  =  2/2.803  = 
0.714;  then  P/a  =  ioooo  pounds  per  square  inch,  cf/r2  =  0.039, 
and  (P /a) (cf/r2)  sech6  =  2jS  pounds  per  square  inch,  which  is 
the  flexural  stress  due  to  the  moment  of  P.  Lastly,  the  total 
tensile  stress  on  the  convex  side  of  the  middle  of  the  eye -bar  is 
5  =  i  240  —  278  + 10  ooo  =  10  960  pounds  per  square  inch. 

Formula  (102)"  also  applies  to  a  beam  uniformly  loaded  and 
under  the  tension  P  by  reversing  the  sign  of  P,  and  thus, 

Si  =  (wcE/P)  (i  -sechfl)  (103)" 

where  6  is  the  number  %(P12/EI)*  when  the  ends  of  the  beam  are 
supported.  For  the  above  eye-bar,  sech0  =0.714;  also  w  = 
490/216  pounds  per  linear  inch.  Then  the  formula  gives  S\  = 
941  for  the  flexural  unit-stress,  so  that  the  total  compressive  unit- 
stress  is  S  =941  + 10  ooo  =  10  941  pounds  per  square  inch. 

The  three  methods  hence  give  numerical  results  which  are 
essentially  the  same  for  all  practical  purposes,  but  the  first  one 
is  the  most  convenient  in  computation  and  hence  is  generally 
preferable.  Formula  (103)  applies  to  all  kinds  of  loading,  and 
to  all  arrangements  of  ends,  as  also  does  (103)';  but  (103)" 
applies  only  to  uniform  load  and  for  this  case  it  is  theoretically 
more  correct  than  the  other  formulas. 

Since  many  students  will  here  meet  with  hyperbolic  functions 
for  the  first  time,  it  may  be  explained  that  they  are  closely  anal- 
ogous to  circular  trigonometric  functions.  For  circular  functions 
cos20  +  sin20  =  i,  but  for  hyperbolic  functions  cosh20-sinh20  =  i. 


ART.  104         ECCENTRIC  AXIAL  FORCES  ON  BEAMS  261 

The  value  of  cos#  is  \(eid  +  e~ie) ,  in  which  e  is  the  Naperian  base 
2.71828  and  i  is  the  square  root  of  —  i;  the  value  of  coshfl  is 
givQn  by  the  simpler  expression  %(ee  +  e~e).  The  reciprocal  of 
cos#  is  seed  and  that  of  coshtf  is  sech#.  Hyperbolic  functions  are 
of  great  importance  in  the  theory  of  electricity  and  in  other  dis- 
cussions of  applied  mechanics;  a  table  of  values  of  such  func- 
tions may  be  found  in  McMahon's  Hyperbolic  Functions 
(Mathematical  Monograph  No.  4,  New  York,  1906). 

Prob.  103a.  A  wooden  cantilever  beam,  3X4X36  inches,  has  a 
load  of  650  pounds  at  the  end  and  is  under  the  longitudinal  compres- 
sion of  4  500  pounds.  Compute  the  maximum  compressive  unit-stress. 

Prob.  1036.  When  the  above  cantilever  beam  is  under  the  longitu- 
dinal tension  of  4  500  pounds,  compute  the  maximum  unit-stress  due 
to  it  and  the  load  of  650  pounds  at  the  end. 

ART.  104.    ECCENTRIC  AXIAL  FORCES  ON  BEAMS 

In  the  three  preceding  articles  the  axial  forces  applied  to  the 
beam  have  been  supposed  to  act  at  the  centers  of  gravity  of  the 
end  sections,  so  that  the  stresses  due  to  them  would  be  uniformly 
distributed  over  every  section  area  were  it  not  for  the  deflection  of 
the  beam.  Sometimes,  however,  these  axial  forces  are  applied 
eccentrically  at  the  ends,  as  shown  in  the  following  figures,  Fig. 
104a  representing  a  compression  applied  through  pins  which  also 
serve  as  supports  for  the  beam,  and  Fig.  1046  representing  a  ten- 
sion applied  in  a  similar  way.  In  the  first  figure  the  longitudinal 
compressive  forces  P  are  applied  below  the  centers  of  gravity  of 
the  end  sections,  this  being  done  in  order  that  the  moment  of  P 
may  tend  to  decrease  the  deflection  of  the  beam  instead  of  in- 
creasing it  as  is  the  case  when  they  are  applied  axially  at  the 
ends.  It  is  required  to  find  the  amount  of  this  eccentricity  so 
that  the  unit-stress  5  at  the  middle  of  the  beam  shall  be  uniform 
and  equal  to  P/a  over  the  entire  cross-section. 

When  the  stress  is  uniform  over  the  section  at  the  middle 
of  the  beam,  there  can  be  no  flexural  stresses  in  that  section  and 
hence  no  bending  moment.  To  insure  this  condition,  it  may 
be  considered,  as  an  approximation,  that  the  moment  of  P  should 


262  APPARENT  COMBINED  STRESSES  CHAP.  XI 

be  equal  to  the  moment  M  of  the  transverse  loads.  Let  p  be 
the  distance  of  P  below  the  center  of  gravity  of  the  end  sections 
in  Fig.  104a  or  above  it  in  Fig.  104ft.  Then  when  Pp=M  there 
is  no  bending  moment  at  the  middle;  accordingly  the  required 
eccentricity  is  p=M/P. 

For  example,  take  the  steel  upper  chord  of  a  bridge  which 
has  a  length  of  30  feet  between  the  pins  at  its  ends.  The  section 
is  made  up  of  two  channels  and  a  plate,  as  in  Fig.  76c,  the  section 
area  being  20.5  square  inches  and  its  moment  of  inertia  742 
inches4.  This  chord  is  subject  to  a  longitudinal  compression 
of  1 68  ooo  pounds,  and  it  is  required  to  find  the  distance  p  below 
the  neutral  axis  at  the  ends  where  the  centers  of  the  pins  should 
be  located.  The  weight  of  the  beam  is  2  090  pounds,  and,  taking 
it  as  supported  at  the  ends,  the  moment  due  to  its  weight  is  M  = 
JTF/  =  94ooo  pound-inches.  Then  the  centers  of  the  pins  must 
be  at  the  distance  ^=94000/168000=0.56  inches  below  the 
axis  of  the  chord  in  order  that  no  flexural  stresses  may  exist. 
The  compressive  stress  over  the  middle  section  is  then  uniform 
and  equal  to  168  000/20.5  =-8  200  pounds  per  square  inch. 


P- 


-  P 


Fig.  104a  Fig.  1046 

The  above  method  is  not  exact,  because  it  takes  no  account 
of  the  stiffness  of  the  beam  and  gives  the  same  results  for  all 
beams  of  the  same  weight.  A  better  method  may  be  derived 
by  considering  the  beam  to  have  the  deflection  /  before  the 
eccentric  load  P  is  applied.  Then  for  compression  p  must  be 
greater  than  /  and  the  moment  P(p-f)  should  equal  if;  for 
tension  the  moment  P(p+f)  is  to  be  equal  to  M.  Accordingly, 

P-j+f        P-j-f  do4) 

the  first  of  which  applies  to  compression  and  the  second  to  ten- 
sion. Hence  values  of  p  computed  from  these  formulas  will 
be  greater  for  compression  and  less  for  tension  than  those  found 
from  the  preceding  method. 


ART.  105  SHEAR  AND  AXIAL  STRESS  263 

Using  the  data  of  the  above  chord  member  while  it  is  under 
the  longitudinal  compression  of  168  ooo  pounds,  the  deflection 
due  to  its  own  weight  is  /=5^/3/384E/  =  o.o57  inches,  and 
M/P  =  o.$6  inches  as  before;  then  the  required  eccentricity  is 
^  =  0.56  +  0.06=0.62  inches.  This  same  section  might  serve 
for  a  lower  chord  under  a  tension  of  168  ooo  pounds,  in  which 
case  the  second  formula  given  ^=0.56  —  0.06=0.50  inches  for 
the  eccentricity.  These  values  are  more  reliable  than  the  eccen- 
tricity 0.56  inches  which  preceding  method  gives  for  both  com- 
pression and  tension. 

Prob.  104a.  Compute  the  eccentricity  p  that  is  required  for  the 
wooden  beam  which  is  discussed  in  Art.  102. 

Prob.  104ft.  Compute  the  eccentricity  p  that  is  required  for  the  eye- 
bar  of  Art.  103  in  order  that  there  may  be  little  or  no  flexural  stress 
at  the  middle. 

ART.  105.     SHEAR  AND  AXIAL  STRESS 

Let  a  bar  having  the  section  area  a  be  subjected  to  the  longi- 
tudinal tension  or  compression  P,  and  at  the  same  time  to  a 
shear  V  at  right  angles  to  its  length.  The  axial  unit-stress  on 
the  section  area  is  P/a,  which  may  be  designated  by  5,  and  the 
shearing  unit-stress  is  V/a,  which  may  be  denoted  by  S8.  It  is 
required  to  find  the  maximum  unit-stresses  produced  by  the 
combination  of  5  and  S8.  In  the  following  demonstration  S 
will  be  regarded  as  a  tensile  unit-stress,  although  the  reasoning 
and  conclusions  apply  equally  well  when  it  is  compressive. , 

Consider  an  elementary  cubic  particle,  with  edges  one  unit 
in  length,  acted  upon  by  the  horizontal  normal  unit-stress  S  and 
by  the  vertical  shearing  unit-stresses  Ss  and  S8,  as  shown  in 
Fig.  105a.  These  forces  are  not  in  equilibrium  unless  a  horizontal 
couple  be  applied  as  in  the  figure,  each  of  the  forces  of  this  couple 
being  equal  to  S8.  Therefore  at  every  point  of  a  body  under 
vertical  shear,  there  exists  a  horizontal  shearing  unit-stress  equal 
to  the  vertical  shearing  unit-stress.  Heretofore  only  one  of  these 
shearing  stresses  has  been  noted,  namely,  that  which  is  parallel 
to  the  applied  external  shear,  but  it  is  now  seen  that  this  is  always 


264 


APPARENT  COMBINED  STRESSES 


CHAP.  XI 


accompanied  by  another  shearing  stress.  For  example,  at  any 
point  in  a  beam  where  there  is  a  vertical  shearing  unit-stress 
V/dj  there  is  also  found  a  horizontal  shearing  unit-stress  of  the 
same  intensity.  Similarly,  Fig.  90  shows  the  shearing  stresses 
rormal  to  the  radius  of  a  shaft  under  torsion,  but  there  are  also 
shearing  stresses  parallel  to  the  radius  which  have  the  opposite 
direction  of  rotation. 


Fig.  105a 


Fig.  1056 


Let  the  parallelopipedal  element  in  Fig.  1056  have  the  length 
dXj  the  height  dy,  the  diagonal  dz,  and  a  width  of  unity  normal 
to  the  plane  of  the  paper.  The  tensile  force  S  .  dy  tends  to  pull 
it  apart  longitudinally.  The  vertical  shear  Ss  .  dy  tends  to  cause 
rotation  and  this  is  resisted  by  the  horizontal  shear  Ss  .  doc.  These 
forces  may  be  resolved  into  components  normal  and  parallel  to 
the  diagonal  dz,  as  shown  in  the  figure.  The  components 
normal  to  the  diagonal  form  normal  tensile  force  Sn  •  dzy  and  those 
parallel  to  the  diagonal  form  a  shearing  force  Sp  .  dz,  where  Sn 
and  Sp  are  the  normal  tensile  and  the  shearing  unit-stresses  upon 
and  along  the  diagonal.  It  is  required  to  find  the  maximum 
values  of  Sn  and  Sp  due  to  the  given  unit-stresses  S  and  S8. 

Let  (j>  denote  the  angle  between  doc  and  dz.  Resolving  each 
of  the  given  forces  in  directions  perpendicular  and  parallel  to 
the  diagonal,  and  taking  their  sum,  there  results, 

Sndz=S  .  dy  .  sin^+Sg  .  doc  .  sm<j)  +  Ss  .  dy  .  cos<£ 
Spdz=S  .  dy  .  cos(j)+Sa  .  dx  cos(j)—Ss  .  dy  sm<f> 
Since  dx=dz.cos<l>  and  dy=*dz.sw$,  these  equations  reduce  to 


By  differentiating  each  of  these  with  respect  to  <f>  and  equating 
each  derivative  to  zero,  it  is  found  that, 


ART.  105  SHEAR  AND  AXIAL  STRESS  265 


Sn  is  a  maximum  or  minimum  when  cot2^>=  — 

Sp  is  a  maximum  when  tan2^>  =  +  %S'/S8 

Expressing  cos>2<f>  and  shi2</>  in  terms  of  cot2<£  and  tan2<£  and 
inserting  their  values  in  the  expressions  for  Sn  and  SP)  the  follow- 
ing important  results  are  obtained: 

max  Sn  =  %S±  (S82  +  (iS)2)*  max  Sp  =  (S2  +  (i^)2)*      (105) 

These  formulas  apply  when  S  is  either  tension  or  compression. 
When  S  is  tension  the  plus  sign  before  the  radical  is  to  be  used 
to  find  the  maximum  tensile  unit-stress  Sn,  while  the  minus  sign 
gives  the  maximum  compressive  unit-stress  Sn- 

For  example,  take  a  bolt  one  inch  in  diameter  which  is  sub- 
ject to-  a  longitudinal  tension  of  5  ooo  pounds  and  at  the  same 
time  to  a  cross-shear  of  3000  pounds.  Here  5  =  6366  pounds 
per  square  inch  and  S8  =  3  820  pounds  per  square  inch.  Then 
Sn  =-1-8  1  55  pounds  per  square  inch,  and  Sn=  —  i  790  pounds 
per  square  inch  for  the  minimum  tensile  or  maximum  compres- 
sive stress,  while  6*^  =  4970  is  the  maximum  shearing  unit-stress; 
these  are  the  greatest  normal  and  shearing  stresses  due  to  the 
combination  of  S  and  S8.  The  directions  which  these  maximum 
stresses  make  with  the  axis  of  the  bolt  are  found  by  using  the 
values  of  cot2^>  and  tan2^>  deduced  above.  For  Sn  the  value  of 
cot2<£  is  -3  183/3820=  -0.833,  whence  0  =  64°  53'  or  <£  =  I54° 
53',  the  former  being  the  inclination  of  the  plane  against  which 
the  maximum  tensile  stress  acts  and  the  latter  being  its  inclination 
for  the  maximum  compressive  stress;  these  two  planes  are  at 
right  angles  to  each  other.  For  Sp  the  value  of  tan2<£  is  +0.833, 
whence  ^  =  19°  53'  or  ^  =  109°  53',  these  being  the  directions  of 
the  planes  along  which  the  maximum  shearing  stresses  act; 
these  directions  bisect  those  of  the  planes  upon  which  the  normal 
stresses  are  the  greatest. 

When  Ss  equals  zero  the  case  is  that  of  simple  tension  or 
compression,  and  max  Sn  =  S,  while  max  SP  =  %S  as  previously 
shown  in  Art.  6.  Here  cot2<£  =  -  oo  or  <£  =  o°;  also  tan2<£  =  oo, 
and  0  =  45°  or  ^  =  135  °  so  that  the  maximum  shearing  stresses 
make  angles  of  45°  with  the  direction  of  the  axial  tension  or 
compression. 


266  APPAKENT  COMBINED  STRESSES  CHAP,  xi 

Prob.  105.  A  bolt  J  inch  in  diameter  is  subjected  to  a  tension  of 
2  ooo  pounds  and  at  the  same  time  to  a  cross  shear  of  3  ooo  pounds. 
Find  the  maximum  tensile  and  shearing  unit-stresses,  and  the  direc- 
tions they  make  with  the  axis  of  the  bolt. 

ART.  106.     COMBINED  FLEXURE  AND  TORSION 

This  case  occurs  when  a  horizontal  shaft  for  the  transmission 
of  power  is  loaded  with  weights.  Let  S  be  the  greatest  flexural 
unit-stress  computed  from  (41)  and  S8  the  torsional  shearing 
unit-stress  computed  from  (90)  or  by  the  special  equations  of  Arts. 
91  and  92.  Then,  according  to  the  last  article,  the  resultant 
maximum  unit-stresses  are, 


max    n  =         \8  max    p= 

the  first  of  whicrf  gives  the  greatest  tensile  or  compressive  unit- 
stress  on  the  lower  or  upper  surface  of  the  shaft,  while  the  second 
gives  the  greatest  shearing  unit-stress.  For  wrought  iron  or 
steel  it  is  usually  necessary  to  regard  only  the  first  of  these  unit- 
stresses,  but  for  timber  the  second  should  also  be  kept  in  view. 

For  example,  let  it  be  required  to  find  the  maximum  unit- 
stresses  for  a  horizontal  steel  shaft,  3  inches  in  diameter  and  12 
feet  between  bearings,  which  transmits  40  horse-power  while 
making  120  revolutions  per  minute,  and  upon  which  a  load  of 
800  pounds  is  brought  by  a  belt  and  pulley  at  the  middle.  Taking 
the  shaft  as  fixed  over  the  bearings,  the  flexure  formula  (41) 
gives  for  the  unit-stress  of  tension  or  compression, 

S=M  .  c/  1  '=  ^Pl/Trd3  =  5  400  pounds  per  square  inch 
From  Art.  92,  the  shearing  unit-stress  on  the  surface  is, 
6^=321  oooH/nd3=4  ooo  pounds  per  square  inch 

The  maximum  tensile  unit-stress  on  the  lower  surface  at  the 
middle  of  the  shaft  or  on  its  upper  surface  in  the  bearings  now  is 

Sn=2  7oo+/\/4ooo2-|-27oo2  =  7  600  pounds  per  square  inch 

and  the  maximum  compressive  unit-stress  on  the  upper  side  of 
the  shaft  has  the  same  value;  this  is  41  percent  greater  than  that 
due  to  pure  flexure.  The  maximum  shearing  unit-stress  is 


ART.  106  COMBINED  FLEXURE  AND  TOKSION  267 

4  900  pounds  per  square  inch,  which  is  22  percent  greater  than 
that  due  to  pure  torsion. 

It  is  thus  seen  that  the  actual  maximum  unit-stresses  in  a 
shaft  due  to  combined  flexure  and  torsion  are  much  higher  than 
those  derived  from  the  formulas  for  flexure  or  torsion  alone. 
In  determining  the  diameter  d  of  a  shaft,  it  is  hence  necessary 
to  take  Sn  as  the  allowable  tensile  or  compressive  stress  and  Sp 
as  the  allowable  shearing  unit-stress.  For  a  round  shaft  of 
diameter  d,  the  expression  for  S  under  any  transverse  load  is 
Mc/I=32M/7rd*  (Arts.  41,  42,  43),  while  that  for  5.  is  Pp  .  c/J  = 
i6Pp/nd3  (Arts.  90,  92).  Inserting  these  in  formula  (105)  and 
solving  for  d3,  there  is  found 


+  M*          T\nd*SP  =     (Pp)2  +  M2    (106) 

in  which  M  is  the  bending  moment  of  the  loa^ds  and  Pp  is  the 
twisting  moment  due  to  the  applied  twisting  forces.  When 
H  horse-powers  are  transmitted  at  a  speed  of  n  revolutions  per 
minute,  the  value  of  Pp  is  given  by  (91).  These  formulas  apply 
only  to  solid  round  shafts;  since  the  allowable  value  of  SP  is 
always  less  than  that  of  Sn,  it  may  often  happen  that  the  second 
formula  will  give  a  larger  diameter  than  the  first. 

As  an  example  let  it  be  required  to  find  the  diameter  of  a 
horizontal  steel  shaft  to  transmit  90  horse-power  at  250  revolu- 
tions per  minute,  when  the  distance  between  bearings  is  8  feet 
and  there  is  a  load  of  480  pounds  at  the  middle,  the  allowable 
unit-stresses  Sn  for  flexure  being  7  ooo  and  that  for  shear  being 
5  ooo  pounds  per  square  inch.  Here  the  bending  moment  M  = 
4X480X96  =  5  760  pound-inches,  and  the  twisting  moment  Pp=* 
63030X90/250  =  22690  pound-inches.  Then  using  the  first 
formula,  the  diameter  d  is  found  to  be  2.8  inches,  while  from  the 
second  formula  it  is  2.9  inches;  hence  the  shaft  should  be  about 
3  inches  in  diameter. 

Formula  (106)  may  also  be  used  for  the  computation  of  the 
maximum  working  unit-stresses  Sn  and  Sp  when  the  shaft  is 
round  and  hollow.  For  a  hollow  shaft  which  has  the  outer  diam- 
eter di  and  the  inner  diameter  d2,  the  formula  will  also  apply  if 
be  substituted  for  d3. 


268  APPARENT  COMBINED  STRESSES  CHAP  XI 

Prob.  1060.  Find  the  factor  of  safety  for  the  data  of  Prob.  92a, 
when  the  shaft  is  in  bearings  12  feet  apart  and  carries  a  load  of  200 
pounds  at  the  middle. 

Prob.  1066.  A  horizontal  nickel-steel  shaft  of  17  inches  outside  and 
ii  inches  inside  diameter  is  to  transmit  16000  horse-powers  at  50 
revolutions  per  minute,  the  distance  between  bearings  being  18  feet. 
Taking  into  account  the  flexure  due  to  the  weight  of  the  shaft,  compute 
the  maximum  unit-stresses. 

ART.  107.     COMPRESSION  AND  TORSION 

When  a  loaded  vertical  shaft  rests  in  a  step  at  its  foot,  the 
torsional  unit-stiess  Ss  combines  with  the  direct  compressive  unit 
stress  S  due  to  the  weight  of  the  shaft  and  its  loads,  and  there  is 
produced  a  resultant  compression  Sn  and  a  resultant  shear  Sp. 
These  may  be  computed  from  (105)  after  S  and  Ss  have  been 
found.  When  W  is  the  load  on  the  section  area  a,  the  value  of  S 
is  W/a  if  the  shaft  is  short,  while  for  long  shafts  it  is  to  be  found 
from  the  column  formula  (80).  In  order  to  prevent  vibration  and 
flexure  it  is  usual  to  .place  bearings  at  frequent  intervals  on  a 
vertical  shaft  so  that  probably  the  use  of  the  column  formula 
will  rarely  be  required,  particularly  when  high  factors  of  safety 
are  used.  The  value  of  S8  is  found  from  the  torsion  formula  (90), 
and  for  a  solid  round  shaft  Ss  =  i6Pp/xd3,  where  Pp  is  the  twist- 
ing moment  which  may  be  replaced  in  terms  of  the  transmitted 
power  by  formula  (92)  . 

To  find  the  diameter  of  a  vertical  solid  round  shaft  for  a  given 
unit-stress  Sn  or  Sp,  a  tentative  method  must  generally  be  em- 
ployed. Inserting  ^W/nd2  for  5  and  i6Pp/xd*  for  Ss  in  for- 
mulas (105),  they  become, 


and  assumed  values  of  d  may  be  substituted  in  each  of  these, 
until  finally  a  value  is  found  which  satisfies  the  equation.  When 
d  is  given,  however,  the  unit-stresses  Sn  and  Sp  may  be  directly 
computed. 

A  vertical  shaft  is  sometimes  so  arranged  that  its  weight  and 
that  of  its  loads  is  supported  near  the  top  on  a  series  of  circular 


ART.  108  HORIZONTAL  SHEAR  IN  BEAMS  269 

disks,  sometimes  called  a  thrust-bearing.  The  shaft  is  thus 
brought  into  tension  instead  of  compression  and  this  is  a  better 
arrangement  because  there  is  then  no  liability  to  lateral  flexure. 
The  above  method  and  formulas  apply  also  to  this  case. 

Prob.  107a.  A  vertical  shaft,  weighing  with  its  loads  6  ooo  pounds, 
is  subjected  to  a  twisting  moment  by  a  force  of  300  pounds  acting  at  a 
distance  of  4  feet  from  its  center.  If  the  shaft  is  wrought  iron,  4  feet 
long  and  2  inches  in  diameter,  find  its  factor  of  safety. 

Prob.  1076.  Find  the  diameter  of  a  short  vertical  steel  shaft  to 
carry  loads  amounting  to  6  ooo  pounds,  when  twisted  by  a  force  of 
300  pounds  acting  at  a  distance  of  4  feet  from  the  center,  taking  the 
working  unit-stress  for  compression  as  10  ooo  and  that  for  shearing  as 
7  ooo  pounds  per  square  inch. 


ART.  108.    HORIZONTAL  SHEAR  IN  BEAMS 

The  common  theory  of  flexure,  as  presented  in  Chapter  V, 
considers  that  the  internal  stresses  at  any  section  are  resolved 
into  their  horizontal  and  vertical  components,  the  former  pro- 
ducing longitudinal  tension  and  compression  and  the  latter  a 
transverse  shear,  and  that  these  act  independently  of  each  other. 
The  shear  formula  Ssa  =  V  supposes  further  that  the  vertical 
shear  is  uniformly  distributed  over  the  cross-section  of  the  beam. 
A  closer  analysis  will  show  that  a  horizontal  shear  exists  also  and 
that  this,  together  with  the  vertical  shear,  varies  in  intensity  from 
the  neutral  surface  to  the  upper  and  lower  sides  of  the  beam.  It 
is  well  known  that  a  pile  of  boards  which  acts  like  a  beam  de- 
flects more  than  a  solid  timber  of  the  same  depth,  and  this  is 
due  to  the  lack  of  horizontal  resistance  between  the  layers.  The 
common  theory  of  flexure  in  neglecting  the  horizontal  shear  gen- 
erally errs  on  the  side  of  safety.  In  some  experiments,  however, 
beams  have  been  known  to  crack  along  the  neutral  surface,  and 
it  is  hence  desirable  to  investigate  the  effect  of  horizontal  shear 
in  tending  to  cause  rupture  of  that  kind.  That  a  horizontal 
shear  exists  simultaneously  with  the  vertical  shear  is  evident 
from  the  considerations  in  Art.  105. 


270  APPARENT  COMBINED  STRESSES  CHAP.  XI 

Let  Fig.  108  represent  a  portion  of  a  beam  of  uniform  section. 
Let  a  notch  nmpq  be  imagined  to  be  cut  into  it,  and  let  forces 

be  applied  to  it  to  preserve 
FT  ,  ~T~    the  equilibrium.     Let  H  be 

m-\—  J-wiT  c 

j  j.  the  sum  of  all  the  horizontal 
components  of  these  forces 
acting  on  mn  and  Hf  the 
sum  of  those  acting  on  qp. 

Now  H'  is  greater  or  less  than  H,  hence  the  difference  Hf  —  H 
must  act  along  mq  as  a  horizontal  shear.  Let  the  distance  mq  be 
ox,  the  thickness  ww'  be  /,  and  the  area  mqq'm'  be  at  a  distance 
z  above  the  neutral  surface.  Let  c  be  the  distance  from  that 
neutral  surface  to  the  remotest  fiber  where  the  unit-stress  is  5. 
Let  da  be  the  section  area  of  any  fiber.  Let  M  be  the  bending 
moment  at  the  section  mn  and  M'  that  at  qp.  Then, 

S/c=  unit-stress  at  distance  unity  from  neutral  surface 
S  .  z/c  —  unit-stress  at  distance  z  from  neutral  surface 
da  .  S  .  z/c=  stress  on  fiber  da  at  distance  z  from  neutral  surface 


of  horizontal  stresses  between  mm'  and  nn' 

Now  from  the  flexure  formula  (41),  S/c  =  M/I  for  the  section  mn 
and  also  S/c  =  M'/I  for  the  section  pq,  where  M  and  M'  are  the 
bending  moments,  and  /  is  the  moment  of  inertia  of  the  entire 
cross-  section.  Accordingly, 


and  hence  the  horizontal  shear  along  mq  is  expressed  by, 
H'-H  =  (M'-M)2czda  .  z/I 

Now,  the  distance  mq  being  3x,  the  difference  M'  —  M  is  dM. 
Also  if  Sh  is  the  horizontal  shearing  unit-stress  on  the  area  /  .  dx, 
the  value  of  H'  '  —  H  is  Shtdx.  Again  from  Art.  47  it  is  known 
that  dM/dx  is  the  vertical  shear  V.  Therefore, 

Sh  =  (V/It)2cda.z  (108) 

is  a  general  formula  for  the  horizontal  shearing  unit-stress  at 
the  distance  z  from  the  neutral  surface  in  any  section  of  a  beam 
where  the  vertical  shear  is  V. 


ART.  108  HORIZONTAL  SHEAR  IN  BEAMS  271 

This  expression  shows  that  the  horizontal  shearing  unit-stress 
is  greatest  at  the  supports,  and  zero  at  the  dangerous  section 
where  V  is  zero.  The  summation  expression  Iczda  .  z  is  the 
statical  moment  of  the  area  mm'nn'  with  reference  to  the  neutral 
axis;  it  is  zero  when  z  =  c,  and  a  maximum  when  z  =  o.  Hence 
the  longitudinal  unit-shear  is  zero  at  the  upper  and  lower  sides 
of  the  beam  and  is  a  maximum  at  the  neutral  surface.  Formula 
(108)  applies  to  any  form  of  section,  /  being  its  width  at  the  dis- 
tance z  from  the  neutral  axis,  and  /  the  moment  of  inertia  of  the 
whole  section  with  respect  to  that  axis.  Since  the  vertical  shear- 
ing unit-stress  at  any  point  is  equal  to  £/„  its  value  at  the  neutral 

surface  is,  y  y 

58=—  I^a  .  z  or  S8=— a^i 

in  which  /  is  the  width  of  the  section  at  the  neutral  axis,  at  is  the 
area  of  the  part  of  the  section  on  one  side  of  the  neutral  axis, 
and  Ci  is  the  distance  of  the  center  of  gravity  of  that  area,  from 
that  axis.  This  formula  gives  the  maximum  shearing  unit-stress 
and  it  is  always  greater  than  the  mean  which  heretofore  has  been 
found  by  dividing  V  by  the  section  area  a. 

For  a  rectangular  beam  of  breadth  b  and  depth  d,  the  value 
of  /  is  b,  and  that  of  /  is  -[lzbd3,  while  the  statical  moment  a\c\ 
is  \bd .  \d>  By  inserting  these  in  the  formula,  there  results 
S*  =  2  V/bd  =  f  .  V/a.  Hence  the  shearing  unit-stress  along  the 
neutral  surface  is  50  percent  greater  than  the  mean  shear  V/a. 

Replacing  /  in  the  above  formula  by  ar2,  where  r  is  the  radius 
of  gyration  of  the  section  area  with  respect  to  the  neutral  axis, 
it  becomes,  y  y 

S>  =  -%-~a  °r  5'=<V 

where  a  is  the  number  aici/tr2,  by  which  the  mean  shear  V/a 
is  to  be  multiplied  in  order  to  obtain  the  maximum  5C  which 
acts  both  horizontally  and  vertically  at  the  neutral  surface.  For 
a  circular  section  of  diameter  d,  the  value  of  /  is  d,  that  of  r2  is 
TV^2>  that  of  a\  is  fad2,  and  that  of  cl  is  2^/3^;  accordingly 
ff  =  f ,  and  the  maximum  shearing  unit-stress  is  Ss  =| .  V/a,  which 
is  33J  percent  greater  than  the  mean. 


272  APPARENT  COMBINED  STRESSES  CHAP.  XI 

For  an  I  section  the  coefficient  a  will  depend  upon  the  ratio 
of  the  flange  and  web  thicknesses  to  their  lengths.  As  a  numerical 
example,  take  a  steel  beam  20  inches  deep  and  weighing  80  pounds 
per  linear  foot;  the  width  of  flanges  being  7.00  inches,  the  mean 
thickness  of  flanges  0.92  inches,  and  the  thickness  of  web  0.60 
inches.  Table  6  gives  ^  =  7.86  inches  and  /  =  o.6o  inches;  the 
statical  moment  a^  is  to  be  found  by  taking  the  sum  of  the 
moments  of  flange  and  web  areas  on  one  side  of  the  neutral  axis 
with  respect  to  that  axis.  The  flange  area  is  7.00X0.92=6.44 
square  inches  and  its  center  of  gravity  is  10.00  —  0.46  =  9.54  inches 
from  the  axis;  the  web  area  on  one  side  of  the  neutral  axis  is 
0.60(10.00  — 0.92)  =5.45  square  inches,  and  its  center  of  gravity 
is  J(io.oo  — 0.92)  =4.54  inches  from  the  axis.  Hence  the  statical 
moment  a\c\  is  6.44X9.54+5.45X4.54  =  86.18  inches3.  The 
coefficient  o  now  is  86. 18/7. 862X  0.6  =  2. 32,  so  that  the  shearing 
unit-stress  at  the  neutral  surface  is  S«  =  2.32F/a  or  132  percent 
greater  than  the  mean  unit-shear  V/a.  It  is  hence  seen  that 
the  shearing  unit-stress  at  the  neutral  surface  of  a  steel  I  beam 
or  plate  girder  should  not  be  computed  by  the  common  formula 
when  a  precise  result  is  required. 

Prob.  108.  In  the  Journal  of  the  Franklin  Institute  for  February, 
1883,  is  described  an  experiment  on  a  spruce  joist  3^X12  inches  and 
14  feet  long,  which  broke  by  tension  at  the  middle  and  afterwards  by 
shearing  along  the  neutral  axis  at  the  end,  when  loaded  at  the  middle 
with  12  545  pounds.  Find  the  maximum  horizontal  shearing  unit- 
stress  by  the  use  of  the  above  formula. 


ART.  109.     LINES  OF  STRESS  IN  BEAMS 

From  the  last  article  it  is  clear  that  at  any  point  in  a  beam 
there  exists  a  horizontal  and  vertical  shearing  unit-stress  S8,  the 
value  of  which  is  given  by  (108).  At  that  point  there  is  also  a 
longitudinal  tensile  or  compressive  unit-stress  S  which  may  be 
computed  from  the  flexure  formula  (41)  with  the  aid  of  the  prin- 
ciple that  these  stresses  vary  directly  as  their  distances  from 
the  neutral  surface.  In  Art.  105  it  was  shown  that  these  unit- 
stresses  combine  to  produce  maxinxum  and  minimum  norma! 


ART.  109  LINES  OF  STRESS  IN  BEAMS  273 

stresses  on  planes  at  right  angles  to  each  other  and  maximum 
shearing  stresses  on  planes  bisecting  these.  The  direction  of 
the  shearing  stresses  and  their  values  are  given  by 


and  from  these  formulas  the  lines  of  maximum  shear  may  be 
traced  throughout  the  beam. 

Art.  108  shows  that  S8  is  greatest  at  the  neutral  surface  and 
zero  at  the  upper  and  lower  sides  of  the  beam.  The  longitudinal 
tensile  or  compressive  stress  S  is  zero  at  the  neutral  surface  and 
greatest  at  the  upper  and  lower  side.  Hence  (f>  is  o  for  the  neutral 
surface,  it  increases  with  the  distance  z  from  that  surface,  and  it 
becomes  45°  at  the  upper  and  lower  sides  of  the  beam.  The 
broken  lines  in  Fig.  109  show  these  lines  of  shear. 

The  angle  6  which  the  direction  of  a  maximum  or  minimum 
normal  stress  makes  with  the  neutral  surface  is  greater  or  less  by 
a  right  angle  than  the  angle  <j>  given  by  cot20  =  —  J5/5«,  because 
<j)  is  the  angle  between  the  neutral  surface  and  the  plane  againsf 
which  Sn  acts.  Accordingly  the  direction  in  which  the  maxi 
mum  tensile  stress  acts  and  its  value  are  given  by, 

Sn 


Now  5  is  the  greatest  at  the  convex  surface  of  the  beam  where  S9 
is  o,  and  hence  6  =  0°.  As  the  neutral  surface  is  approached  5 
decreases  and  Ss  increases,  whence  6  also  increases.  At  the  neu- 
tral surface  S  is  o  and  S8  has  its  greatest  value;  hence  for  that 
surface  0  is  45°.  The  same  conclusions  follow  for  the  maximum 
compressive  stress  on  the  other  side  of  the  neutral  surface. 

The  following  figure  is  an  attempt  to  represent  the  lines  which 
indicate  the  directions  of  the  maximum  unit-stresses  in  a  beam. 
The  full  lines  show  the  directions  of  the  maximum  compressions 
and  the  broken  lines  those  of  the  maximum  tensions,  while  the 
dotted  lines  give  the  directions  of  the  maximum  shears.  On 
any  line  the  intensity  of  stress  varies  with  the  inclination,  being 
greatest  where  the  line  is  horizontal.  The  lines  of  maximum 
shear  cut  those  of  maximum  tension  and  compression  at  angles 
of  45°.  The  lines  of  maximum  tension  and  those  of  maximum 


274 


APPARENT  COMBINED  STRESSES 


CHAP.  Xl 


compression  are  seen  to  cut  each  other  at  right  angles  and 
become  vertical  at  the  lower  and  upper  sides  of  the  beam 
where  those  stresses  are  zero.  There  is  also  another  set 
of  shear  curves  which  cut  at  right  angles  those  shown  in 
the  figure. 


Fig.  109 

It  appears  from  this  investigation  that  the  common  theory  of 
flexure  gives  the  horizontal  unit-stress  correctly  at  the  dangerous 
section  of  a  simple  beam  under  uniform  load,  since  there  the  ver- 
tical shear  is  zero.  At  other  sections  the  stress  S  as  computed 
from  the  flexure  formula  is  correct  for  the  remotest  fiber,  but  for 
other  fibers  there  are  greater  normal  unit-stresses  than  the  com- 
mon theory  gives.  For  a  heavy  concentrated  load,  where  the 
vertical  shear  suddenly  changes  sign  at  the  dangerous  section,  the 
common  theory  gives  the  horizontal  stress  S  correctly  for  the  re- 
motest fiber  only,  and  it  may  be  possible  in  some  forms  of  cross- 
sections  that  this  is  slightly  less  than  the  maximum  Sn  at  a  point 
nearer  to  the  neutral  surface.  This,  however,  is  a  possibility  of 
rare  occurrence,  and  all  that  has  here  been  deduced  justifies  the 
validity  of  the  common  theory  of  flexure  as  a  correct  guide  in  the 
practical  investigation  and  design  of  beams. 

The  resultant  combined  stresses  found  in  this  chapter  are 
called  maximum  'apparent'  stresses,  since  they  are  the  stresses 
which  are  apparently  correct  according  to  the  principles  of  statics. 
It  will  be  seen  later  in  Chapter  XV  that  the  stresses  as  measured 
by  the  deformations  which  occur  must  be  considered,  these  being 
called  'true  stresses'. 

Prob.  109a.  A  joist  fixed  at  both  ends  is  3X12  inches  and  12  feet 
long,  and  is  stressed  by  a  load  at  the  middle,  so  that  the  value  of  S 
as  computed  from  (41)  is  4  ooo  pounds  per  square  inch.  Find  tha 


ART.  109  LINES  OF  STRESS  IN  BEAMS  275 

values  of  Sn  for  points  over  the  support  distant  3,  4,  and  5  inches  from 
the  neutral  surface. 

Prob.  1096.  Show,  for  a  point  between  the  neutral  surface  and  the 
convex  side,  that  there  exists  a  maximum  compression  as  well  as  a 
maximum  tension.  Deduce  an  expression  for  the  value  of  this  maxi- 
mum compression  and  its  direction.  Draw  a  figure  showing  the  curves 
of  maximum  compression  on  both  sides  of  the  neutral  surface  of  a 
cantilever  beam. 

Prob.  109c.  Consult  Weyrauch's  Theorie  der  Trager  (Leipzig, 
1880),  and  examine  his  figures  showing  the  lines  of  stress  in  a  beam 
with  overhanging  end.  Draw  similar  figures  for  a  beam  fixed  at  one 
end  and  supported  at  the  other. 

Prob.  109d.  Consult  Winkler's  Elasticitat  und  Festigkeit  (Prag, 
1867),  and  examine  his  formulas  for  deflection  and  stress  of  a  beam 
under  eccentric  and  axial  tension.  Show  that  these  formulas  may  be 
much  simplified  by  introducing  hyperbolic  functions. 


276 


COMPOUND  COLUMNS  AND  BEAMS 


CHAP.  XII 


CHAPTER    XII 

COMPOUND     COLUMNS    AND    BEAMS 
ART.  110.     BARS  OF  DIFFERENT  MATERIALS 

Heretofore  when  a  bar  or  beam  has  been  mentioned,  it  has 
been  understood  that  it  was  of  the  same  material  throughout. 
It  is,  however,  possible  to  have  a  bar  in  which  different  materials 
are  combined,  and  beams  of  this  kind  are  now  common  in  con- 
crete-steel construction.  Timber  and  steel  are  also  sometimes 
combined  in  one  beam,  especially  for  the  floor  stringers  of  elec- 
tric railway  bridges. 


p 

1 

i 

LJ 

1 

2 

2 

3 

1         => 

Fig.  llOa  Fig.  1106 

In  the  case  of  a  bar  the  different  materials  might  occupy  dif- 
ferent parts  of  the  length  as  shown  in  Fig.  llOa,  where  the  spaces 
i,  2,  3  designate  different  materials,  although  this  is  a  method 
rarely  used.  In  Fig.  1106  the  three  materials  are  shown  arranged 
longitudinally,  this  being  the  method  most  commonly  employed 
for  compound  bars.  Round  bars  for  jail  windows  are  sometimes 
made  of  two  kinds  of  steel,  an  inner  core  of  soft  steel  and  an 
outer  annulus  of  very  hard  steel,  the  function  of  the  former  being 
to  resist  lateral  bending  and  that  of  the  latter  to  resist  attempts  of 
the  prisoners  to  file  or  cut.  A  compound  bar  may  also  be  formed 
of  timber  and  plates  of  metal  bolted  together,  this  being  a  method 
more  commonly  used  for  beams  than  for  bars.  In  these  figures 
the  bars  are  represented  as  short  columns  under  compression, 
but  the  following  reasoning  is  general  and  applies  also  when  com- 
pound bars  are  under  tension. 


ART.  no  BARS  OF  DIFFERENT  MATERIALS  277 

The  load  P  on  the  bar  is  supposed  to  be  axial,  so  that  the 
stress  is  uniformly  distributed  over  the  section  area  of  each 
material.  Considering  first  the  case  of  Fig.  1100,  where  the 
materials  are  arranged  in  series,  it  is  plain  that  the  total  stress 
in  each  section  area  is  equal  to  P.  Let  lit  12,  /3  be  the 
lengthr;  of  the  three  parts,  #1,  a2,  a3  their  section  areas,  and 
EI,  E2f  E3  the  moduluses  of  elasticity  of  the  three  materials. 
The  change  of  length  e\  of  the  first  part  is  Pli/aiEi,  that  of  the 
second  part  is  ea  =  P12/  'a2E2,  and  that  of  the  third  part  is  Pl3/a3E3 
(Art.  10).  Accordingly, 


E3a3' 

is  the  total  change  in  length  of  the  bar  provided  the  elastic  limit 
of  the  material  be  not  exceeded.  The  unit-stresses  in  the  three 
parts  of  the  bar  are  P/0i,  P/a2,  P/a3;  when  the  three  section 
areas  are  equal,  as  in  the  first  diagram  of  Fig.  11  0#,  the  unit- 
stresses  in  the  three  parts  are  equal. 

For  the  case  of  Fig.  1106,  the  three  lengths  are  equal  to  /, 
and  the  three  section  areas  are  0i,  a2,  a3.  Let  PI,  P2,  P3  denote 
the  loads  borne  by  the  parts,  i,  2,  3,  the  sum  of  these  being  equal 
to  P.  These  loads  may  be  found  from  the  fact  that  the  changes 
of  length  of  the  three  parts  must  be  equal.  Accordingly, 
P!  -|-  P2  +  P3  =  P  Pi/AiEi  =  P2l/a2E2  P3l/a3E3  =  P2l/<i2E2 

are  three  necessary  conditions,  and  their  solution  gives, 
Pi=P.aiEi/D  P2=P.a2E2/D  P3=P  .  a3E3/D     (110)' 

in  which  D  represents  the  quantity  a\E\  +  a2E2+a3E3.  After 
PI,  P2,  P3  have  been  computed  from  these  formulas,  the  unit- 
stresses  in  the  three  parts  of  the  bar  are  found  from  S\=P\/ai, 
S2=P2/a2,  S3  =  P3/a3.  These  conclusions  are  not  valid  unless 
each  of  these  unit-stresses  is  less  than  the  elastic  limit  of  the 
material,  because  the  formulas  for  change  of  length  apply  only 
under  this  condition.  By  substituting  either  of  the  values  of  P 
in  the  corresponding  expression  for  change  of  length,  there  is 


which  differs  materially  from  (110).     These  two  formulas  have 


278  COMPOUND  COLUMNS  AND  BEAMS  CHAP.  XII 

a  marked  analogy  with  the  electric  equations  which  connect  loss 
of  voltage  with  current  in  wires  laid  in  series  and  in  parallel, 
and  this  analogy  will  be  discussed  in  Art.  185. 

As  a  numerical  example  illustrating  the  case  of  Fig.  llOa, 
let  the  three  materials  be  timber,  stone,  and  steel,  so  that  E\  = 
i  500  ooo,  E2  =  6  ooo  ooo,  £3  =  30  ooo  ooo  pounds  per  square  inch. 
Let  each  section  area  be  a  and  each  length  be  J/.  Then  the 
unit-stress  in  every  section  area  is  P/a,  and  formula  (110)  gives 
the  total  change  of  length  as  e  =  0.000  ooo  2&C)Pl/a. 

For  the  case  of  Fig.  110ft,  let  the  three  materials  be  also  timber, 
stone,  and  steel;  let  each  length  be  /  and  each  section  area  be  Ja. 
Then  formulas  (110)'  give  PI  =  o.o4P,  P2  =  o.i6P,  and  P3  =  o.8oP, 
so  that  the  stiffest  material  carries  the  greatest  load,  and  the 
unit-stress  in  the  steel  is  twenty  times  that  in  the  timber.  From 
(110)"  the  change  of  length  is  e  =  o.ooo  ooo  o~oP//a,  or  less 
than  one-third  of  that  of  the  previous  case.  These  last  two 
columns  contain  the  same  amount  of  each  material;  but  the 
total  change  of  length  is  the  greatest  for  the  first  case,  while  the 
unit-stress  in  the  steel  is  the  greatest  for  the  second  case. 

As  a  numerical  example  illustrating  the  second  diagram  of 
Fig.  110ft,  let  the  central  part  be  of  timber,  6x8  inches  in  sec- 
tion, and  let  two  steel  plates,  each  f  inches  thick  and  8  inches 
wide,  be  bolted  to  the  8-inch  sides.  Let  the  length  of  the  short 
column  be  5  feet,  and  the  axial  load  P  be  126  ooo  pounds.  Here 
/  =  6o  inches,  01=^3  =  3  square  inches,  #2=48  square  inches, 
EI  =Ea  =  30  ooo  ooo  and  £2  =  1  500  ooo  pounds  per  square  inch. 
Then  the  formulas  give  PI  =  PS  =  -ffP  =  45  ooo  pounds,  and 
P2  =  fP  =  36  ooo  pounds.  The  compressive  unit-stress  in  the  steel 
plates  then  is  51=45000/3  =  15000,  and  that  in  the  timber  is 
£2  =  36  000/48  =  750  pounds  per  square  inch.  In  this  case  the 
two  steel  plates  carry  about  70  percent  of  the  total  load,  and  are 
so  highly  stressed  that  bolts  should  be  placed  at  frequent  intervals 
in  order  to  prevent  lateral  buckling. 

Steel  ropes  are  often  made  with  a  hemp  core  in  order  to  give 
flexibility,  and  here  also  the  tensile  load  is  divided  between  the 
two  materials  inversely  as  their  resistances.  Thus,  if  0i  and  a2 


ART.  ill  REINFORCED  CONCRETE  COLUMNS  279 

are  the  section  areas  of  the  hemp  and  steel,  and  E\  and  E2  their 
moduluses  of  elasticity,  then  Pi/P2  =  a\Ei/a2E2)  which  shows 
that  the  steel  takes  nearly  all  the  load,  since  a2  is  usually  equal 
to  60i,  while  E2  is  probably  more  than  100  times  as  great  as  E\. 

Prob.  110,  A  bar  for  a  jail  window  has  a  diameter  of  2\  inches,  the 
central  core  of  soft  steel  being  ij  inches  in  diameter.  When  this  bar 
is  used  under  axial  tension  or  compression,  show  that  the  unit-stress 
on  the  hard  and  soft  steel  is  the  same. 


ART.  111.    REINFORCED  CONCRETE  COLUMNS 

A  common  instance  of  a  compound  column  is  that  shown  in 
section  in  Fig.  Ilia,  which  represents  a  hollow  cylinder  of  cast 
iron  or  steel  filled  with  concrete  and  used  for  one  of  the  supports 
of  a  bridge.  The  usual  intention  is  that  the  concrete  shall  carry 
the  load,  while  the  metallic  cylinder  is  to  prevent  the  concrete 
from  cracking  under  the  action  of  the  weather  or  of  collisions 
from  floating  objects.  Owing  to  the  friction  between  the  two 
materials,  however,  it  is  evident  that  the  metal  always  carries 
part  of  the  weight,  and  the  highest  load  that  can  come  upon  it 
is  readily  found  from  the  method  of  Art.  110.  Let  P  be  the 
total  load  on  the  pier  or  column,  a\  and  a2  the  section  areas  of 
the  concrete  and  metal,  and  EI  and  E2  the  moduluses  of  elasticity 
of  the  same;  then  (110)'  reduce  to, 


are  the  loads  which  come  on  concrete  and  metal  respectively. 
For  example,  take  a  pier  where  the  concrete  is  6  feet  in  diameter, 
this  being  surrounded  by  a  cast-iron  casing  1.15  inches  thick. 
Using  the  average  values  in  Table  2,  the  ratio  E2/Ei  is  6  ;  from 
Table  16  the  area  a\  is  4071  square  inches,  while  the  area  a2 
is  with  sufficient  precision  n  X  73  X  1.15  =  265  square  inches,  and 
hence  the  ratio  ai/a2  is  15.4.  Then  the  formulas  give  Pi=o.72P 
and  P2  =  o.28P,  so  that  the  cast  iron  may  carry  about  one-fourth 
of  the  load. 

The  unit-stresses  in  the  different  materials  of  a  compound 


280  COMPOUND  COLUMNS  AND  BEAMS  CHAP.  XI* 

column  are  proportional  to  their  moduluses  of  elasticity,  that  is, 
to  the  stiffnesses  of  the  materials.  This  is  readily  seen  from  (110)', 
or,  for  the  case  of  two  materials, 

Pil/aiEi  =  P2l/a2E2     or     S2/Sl  =  E2/El 

For  example,  referring  to  the  column  of  the  last  paragraph,  the 
value  of  E2/Ei  is  5,  and  hence  the  unit-stress  52  in  the  cast-iron 
is  5  times  the  unit-stress  Si  in  the  concrete. 

When  bars  or  rods  of  metal  are  placed  in  a  concrete  column 
it  is  said  to  be  'reinforced'.  Square  or  rectangular  columns 
are  built  with  vertical  steel  rods  located  near  the  corners,  as  seen 
in  Fig.  1116.  In  construction  the  rods  are  first  put  in  position, 
they  being  connected  by  heavy  horizontal  wires  in  order  to  keep 
them  in  place  while  the  concrete  is  packed  in  a  wooden  form 
which  is  built  around  them.  Some  of  the  columns  of  the  Harvard 
stadium  are  14  inches  square  and  have  vertical  steel  rods  f  inches: 
in  diameter  near  the  corners.  Here  the  section  area  of  tht 
four  steel  rods  is  a2  =0.442  square  inches,  while  that  of  the  concrete 
is  #i  =  196  —  0.44  =  195.56  square  inches,  so  that  the  ratio  a^/a2  = 
442.  The  average  value  of  the  ratio  E2/Ei  is  about  10.  Accord- 
ingly, the  formulas  give  for  the  load  on  the  concrete  PI  =0.978? 
and  for  the  load  on  the  steel  P2=o.o22P.  The  allowable  unit- 
stress  for  the  concrete  was  taken  as  350  pounds  per  square  inch, 
hence  it  follows  from  the  last  paragraph  that  the  unit-stress  in 
the  steel  rods  is  about  3  500  pounds  per  square  inch.  Here  the 
full  strength  of  the  metal  is  not  developed,  and  this  is  usually 
the  case  in  reinforced  concrete  construction. 

In  the  above  discussion  the  column  is  regarded  as  so  short 
that  no  account  need  be  taken  of  the  lateral  flexure,  and  this 
may  be  safely  done  until  the  height  of  the  column  exceeds  about 
twelve  times  its  least  diameter.  Thus  a  column  14  inches  square 
may  be  as  high  as  14  feet  before  it  is  necessary  to  use  any  of  the 
formulas  given  for  columns  in  Chapter  IX;  the  slenderness-ratio 
l/r  corresponding  to  this  rule  is  about  40  for  a  square  column 
and  48  for  a  round  column.  For  higher  values  of  l/r,  the  method 
of  investigation  for  the  concrete  part  is  to  find  the  load  PI  as 
above,  and  then  compute  the  unit-stress  Si  from  the  column 


ART.  Ill 


REINFORCED  CONCRETE  COLUMNS 


281 


formula  (80),  using  for  the  factor  <£  a  high  number,  say  about 
-nnnr>  because  sufficient  experiments  have  not  been  made  to 
determine  its  proper  value.  As  for  the  steel,  it  is  everywhere 
supported  by  the  concrete  and  can  have  no  lateral  flexure  except 
that  due  to  the  concrete  part,  but  it  will  be  fair  to  consider  that 
its  unit-stress  S2  is  increased  in  the  same  proportion  as  Si,  due 
regard  being  paid  to  its  distance  from  the  axis  of  the  column. 
For  instance,  let  Si  and  S2  as  computed  for  a  short  prism  be 
300  and  3  ooo  pounds  per  square  inch,  let  Si  on  the  concave 
side  of  the  column  be  found  by  Rankine's  formula  to  be  500; 
then  S2  will  be  5  ooo  if  the  steel  is  located  close  to  the  concave 
side,  but  if  it  be  half-way  between  that  side  and  the  axis  S2  will 
be  only  3000  +  ^X10X200=4000  pounds  per  square  inch. 
In  general,  the  unit-stress  in  the  concrete  reaches  its  allowable 
limit  before  the  steel  receives  a  stress  of  one-half  that  which  is 
permissible. 


Fig.  Ilia 


Fig.  lllc 


A  concrete  post  is  often  made  having  a  steel  rod  running 
longitudinally  through  it  at  the  axis  as  in  Fig.  lllc,  this  being 
for  the  purpose  of  resisting  lateral  flexure  rather  than  to  assist  in 
carrying  loads.  Concrete  piles  are  made  on  the  same  plan  as 
the  bridge  column  above  described,  the  concrete  being  enclosed 
in  a  metal  cylinder.  The  tunnel  under  the  Hudson  river,  under 
construction  by  the  Pennsylvania  Railroad  in  1906,  is  supported 
on  steel  screw  piles  filled  with  concrete.  Foundation  walls  of 
concrete  sometimes  have  vertical  steel  rods  which  help  to  carry 
a  part  of  the  weight  and  at  the  same  time  prevent  the  concrete 
from  cracking.  Reinforced  concrete  beams  are  discussed  in 
Arts.  113-116.  Concrete  pipes,  sewers,  arches  for  buildings  and 
for  bridges,  are  built  in  which  the  steel  reinforcement  plays  a 
more  important  part  than  it  does  in  columns.  Nearly  all  of  this 


282 


COMPOUND  COLUMNS  AND  BEAMS 


CHAP.  XII 


reinforced  concrete  steel  construction  has  been  developed  since 
the  first  edition  of  this  book  was  published. 

Prob.  111.  A  concrete  pile,  as  designed  for  the  foundation  of  a 
building  in  New  York,  was  12X12  inches  in  section  area,  and  had 
four  vertical  steel  rods,  each  i-^  inches  in  diameter  and  placed  at 
about  two  inches  from  the  corner.  Compute  the  unit-stresses  in  con- 
crete and  steel  at  a  depth  of  12  feet  below  the  top,  due  to  their  own 
weight  and  to  a  load  of  30  ooo  pounds  on  the  top. 


ART.  112.    FLITCHED  BEAMS 

A  'flitched  beam'  is  one  made  of  timber  with  metal  plates 
upon  its  sides,  these  being  held  in  place  by  bolts  passing  through 
the  timber;  sometimes,  however,  a  single  plate  is  placed  between 
two  timber  beams.  The  following  figures  show  sections  of  such 
beams;  the  third  one,  which  is  formed  by  two  channels  and  a 
piece  of  timber,  is  often  used  on  highway  bridges  for  stringers 
which  support  tracks  of  an  electric  railway,  the  rail  being  fastened 
to  the  timber  part  by  spikes  or  lag  bolts.  The  other  forms  are 
sometimes  employed  in  wooden  floors.  The  bolts  should  in  all 
cases  pass  through  the  neutral  axis  of  the  section,  in  order  to 
weaken  the  beam  as  little  as  possible. 


Fig.  112a 


Fig.  1126 


Fig.  112c 


When  a  load  is  placed  upon  such  a  beam  it  divides  itself  be- 
tween the  two  materials  in  proportions  depending  upon  their 
Stiffness  and  section  areas.  Whether  the  load  be  concentrated  or 
uniform  it  may  be  expressed  by  W,  and  this  will  divide  into  two 
parts,  W\  being  that  carried  by  the  timber  and  W^  that  carried 
by  the  metal.  Since  the  two  materials  are  fastened  together  the 
deflection  of  each  is  the  same.  These  conditions  enable  the  values 
of  W\  and  W^  to  be  determined  in  a  manner  similar  to  that  used 
for  the  compound  column  in  Art.  110.  The  length  of  the  metal 


ART.  112  FLITCHED  BEAMS  283 

and  timber  will  be  taken  the  same,  each  being  equal  to  the  length 
/  of  the  span  of  the  beam.  The  modulus  of  elasticity  of  the  tim- 
ber will  be  denoted  by  EI  and  that  of  the  steel  by  E2  ;  the  moment 
of  inertia  of  the  cross-section  of  the  timber  is  I\  and  that  of  the 
metal  is  I2. 

The  deflection  /  of  a  beam  may  be  expressed,  as  in  Art.  56, 
by  Wl3/£tEI,  where  /?  is  a  number  depending  upon  the  kind  of 
loading  and  the  arrangement  of  the  ends.  From  the  above  con- 
ditions, 


and  the  solution  of  these  equations  gives  the  values, 


which  are  seen  to  be  the  same  as  (111)  except  that  a\  and  a2  are 
replaced  by  I\  and  I2.  After  the  values  of  W\  and  W2  have  been 
computed,  the  unit-stresses  in  the  timber  and  steel  may  be  inves- 
tigated by  the  method  of  Art.  48. 

For  example,  let  a  flitched  timber  beam  like  Fig.  1120  be  8X  12 
inches  in  section  and  each  of  the  steel  plates  be  JX9  inches.  For 
the  timber  /i  =  •}•$  X  8X  1  23  and  for  the  steel  I2  =^2  X  i  X  Q3,  whence 
the  ratio  /i//2  =  512/27;  also  the  ratio  E2/Ei=2o.  '  Then  (112) 
gives  W\  =0.487^  and  ^2  =  0.513^,  so  that  the  parts  of  the  load 
carried  by  timber  and  steel  are  closely  equal.  Let  the  length  of 
this  simple  beam  be  15  feet  and  the  total  uniform  load  on  it  be 
W  =  16000  pounds,  so  that  Wi=j8oo  and  W2  =  8  200  pounds. 
From  the  flexure  formula  (41)  the  unit-stress  on  the  upper  and 
lower  sides  of  the  timber  at  the  middle  of  the  span  is  Si  =M\c\II\  = 
J^i/Ci/8/i  =914  pounds  per  square  inch;  also  the  unit-stress  on 
the  upper  and  lower  sides  of  the  steel  at  the  same  section  is  £2  = 
M2c2/I2  =  W2lc2/8I2  =  13  700  pounds  per  square  inch.  These 
unit-stresses  are  safe  allowable  values  for  the  conditions  under 
which  such  a  beam  would  generally  be  used. 

To  design  a  flitched  beam,  the  size  of  the  timber  is  first  as- 
sumed and  then  the  proper  thickness  and  depth  of  the  metal 
plates  are  to  be  computed.  Let  Si  and  6*2  be  the  given  allowable 
unit-stresses;  from  the  above  their  ratio  is  Si/S2  =  WiCiI2/W2c2Ii, 


284  COMPOUND  COLUMNS  AND  BEAMS          CHAP.  XII 

and  replacing  W\  and  W2  by  their  values  from  (112)  this  reduces 
to  Si/S2  =  EiCi/E2C2'  The  total  bending  moment  M,  for  any 
kind  of  loading,  is  equal  to  the  sum  of  the  resisting  moments 
Si  .  Ii/ci  and  52  .  1  2/^2-  Accordingly  the  two  equations  for 
design  are, 

Si  .  Ii/ci  +  S2  .  I2/c2=M          (112)' 


in  which  c\  and  c2  are  the  half  -depths  of  timber  and  metal,  and 
hence  the  ratio  d2/di  equals  c2/ci.  For  cast  iron  and  timber  the 
ratio  E2/Ei  is  10,  while  S2/Si  may  be  taken  as  about  4  for  the 
tensile  side  of  the  beam;  hence  the  depth  of  the  metal  should  be 
about  four-tenths  of  the  depth  of  the  timber.  For  structural  steel 
and  timber  the  ratio  E2/E\  is  20,  while  S2/Si  should  be  about 
15;  hence  the  depth  of  the  steel  should  be  three-fourths  of  the 
depth  of  the  timber.  When  di  and  d2  are  equal,  as  in  Fig.  112c, 
the  ratio  Si/S2  equals  Ei/E2,  so  that  the  unit-stress  on  the  tim- 
ber is  one-twentieth  of  that  on  the  steel. 

The  proper  thickness  of  the  metal  plates  will  depend  upon  the 
bending  moment  M  .  For  rectangular  sections  the  moment  equa- 
tion becomes  Sibidi2  +  S2b2d22=6M.  When  the  depths  d\  and 
d2  are  given,  as  also  the  width  bi  of  the  timber,  the  thickness  b% 
of  the  metal  is  computed  from  this  equation.  For  example,  let 
^1  =  12  and  d2  =  g  inches  for  timber  and  steel;  then  S2/Si  must 
equal  15.  Hence  let  Si  =  i  ooo  and  52  =  15  coo  pounds  per  square 
inch  ;  let  61=8  inches,  and  let  it  be  required  to  find  b2  when  the 
total  load  W=i$  ooo  pounds  and  is  concentrated  at  the  middle 
of  a  span  of  16  feet.  Here  M  =  iPF/  =  72oooo  pound-inches,  and 
then  the  moment  formula  gives  b2  =  2%  inches  as  the  total  thick- 
ness of  metal  required. 

In  the  case  of  the  trolley  stringer  of  Fig.  112c,  the  steel  chan- 
nels carry  a  large  part  of  the  total  load  W,  and  they  are  sometimes 
designed  so  that  they  may  carry  it  all.  When  the  depth  of  the 
channels  and  timber  are  the  same,  the  above  theory  shows  that  the 
ratio  Si/S2  equals  Ei/E2,  and  hence  the  flexural  unit-stress  on 
the  channels  is  twenty  times  that  on  the  timber.  Thus,  if  £2  for 
the  structural  steel  is  taken  as  12  coo  pounds  per  square  inch,  Si 
for  the  timber  will  be  600  pounds  per  square  inch.  For  example, 


ART.  113  REINFORCED  CONCRETE  BEAMS  285 

let  the  length  of  a  stringer  be  21  feet,  and  the  total  uniform  load 
upon  it  be  14  ooo  pounds,  this  being  an  equivalent  for  the  total 
live  and  dead  load.  The  moment  M  is  then  %Wl  =  44i  ooo  inch- 
pounds.  Let  the  timber  be  8  inches  wide  and  10  inches  deep; 
then  71/^1  =  133.3  inches3.  Using  Si  =600  and  52  =  i2ooo,  the 
value  of  /2/C2  f°r  the  two  channels  is  found  from  the  moment 
formula  of  (112)'  to  be  30.1  inches3,  whence  from  Table  9  it  is 
seen  that  the  lo-inch  channel  which  weighs  20  pounds  per  foot  is 
required.  Two  channels  of  this  kind  have  /2  =  i57-4  inches4, 
while  for  the  timber  /i  =666.7  inches4.  Accordingly,  by  the  help 
of  (112)  it  is  found  that  the  channels  carry  about  83  percent  of 
the  total  load. 

If  the  metal  plates  shown  in  Figs.  1120  and  1126  do  not 
extends  to  the  ends  of  the  simple  beam,  but  stop  a  distance  id 
from  the  ends,  the  above  method  needs  modification.  By  an 
investigation  to  be  given  in  Art.  123  it  will  be  shown  that  formulas 
(112)  apply  to  this  case  if  /i  is  multiplied  by  i-8/c3.  Hence 
the  shortening  of  the  plates  throws  a  larger  proportion  of  the 
load  upon  the  timber  and  increases  the  stress  in  it  at  the  middle 
of  the  span.  The  condition  Si/S2=EiCi/E2C2  is  also  to  be 
modified  by  multiplying  the  second  member  by  i  —  8/c3,  so  that 
the  advantageous  ratio  c2/Ci  is  less  than  before.  This  case  need 
not  be  discussed  further,  because  plates  extending  over  only  a 
part  of  the  length  would  rarely  be  used  except  in  order  to  strengthen 
a  weak  beam,  and  in  such  an  event  no  precise  computations 
would  be  needed. 

Prob.  112.  Let  a  flitched  beam,  like  Fig.  1126,  consist  of  two  tim- 
bers each  10  inches  wide  and  14  inches  deep,  and  a  steel  plate  J  inches 
thick  and  7  inches  wide.  When  the  unit-stress  in  the  timber  is  900 
pounds  per  square  inch,  what  is  the  unit-stress  in  the  steel?  What 
percentage  does  the  metal  add  to  the  strength  of  the  wooden  beam  ? 

ART.  113.    REINFORCED  CONCRETE  BEAMS 

Other  methods  of  longitudinal  reinforcement  than  that  de- 
scribed in  the  last  article  are  used  for  concrete  beams.  The 
method  of  Fig.  1130  is  occasionally  used,  but  here  the  resistance 


286 


COMPOUND  COLUMNS  AND  BEAMS 


CHAP.  XII 


of  the  concrete  to  flexure  is  not  generally  taken  into  account, 
its  office  being  to  protect  the  steel  beam  from  corrosion  or  fire, 
while  the  steel  beam  is  considered  to  carry  all  the  load.  When 
computations  are  made  on  a  case  of  this  kind,  the  formulas  of 
the  last  article  directly  apply,  the  ratio  E%/E\  being  taken  from 
10  to  15,  while  the  ratio  S%/S\  of  the  allowable  unit-stresses  for 
the  compressive  side  of  the  beam  will  generally  range  from  8  to  12. 
Since  the  ultimate  tensile  strength  of  concrete  is  from  300  to 
500  pounds  per  square  inch,  an  average  allowable  tensile  unit- 
stress  is  about  40  pounds  per  square  inch,  while  that  for  the  steel 
may  easily  under  steady  loads  be  as  high  as  15  ooo  pounds  per 
square  inch.  Supposing  the  steel  to  be  stressed  only  to  8  ooo 
pounds  per  square  inch,  it  is  seen  that  S\  for  the  concrete  will 
be  higher  than  its  tensile  strength,  so  that  the  practice  of  designing 
the  steel  beam  to  carry  all  the  load  is  justified,  the  concrete 
being  considered  only  as  a  covering  which  protects  the  steel  in 
spite  of  the  cracks  on  the  tensile  side. 


II 


Fig.  113a 


Fig.  1136 


Fig.  113c 


Fig.  11 3d 


In  the  method  of  reinforcement  seen  in  Fig.  1136,  there  are 
four  steel  rods  arranged  symmetrically  with  respect  to  the  neutral 
axis;  for  wide  beams  a  larger  number  of  rods  is  used,  half  of 
them  above  and  half  below  the  neutral  axis,  the  distance  of  each 
row  from  that  axis  being  the  same.  The  formulas  of  Art.  112 
apply  directly  to  this  case  when  c\  is  the  half-depth  of  the  rect- 
angular section  and  c2  is  the  distance  from  the  neutral  axis  to 
the  remotest  part  of  the  metal.  When  the  area  of  metal  is  small 
compared  to  that  of  the  concrete,  as  is  generally  the  case,  it  will 
be  sufficiently  precise  to  take  the  moment  of  inertia  1%  as  equal 
to  the  area  a  of  the  metal  multiplied  by  the  square  of  the  distance 
h  from  its  center  of  gravity  to  the  neutral  axis,  or  I2=ah2;  also 
it  will  be  sufficiently  precise  to  take  /i=TVW3,  thus  supposing 
that  the  concrete  fills  also  the  spaces  occupied  by  the  metal. 


ART.  113  REINFORCED  CONCRETE  BEAMS  287 

Considering  the  uncertainties  regarding  the  values  of  the  ratios 
Ez/Ei  and  S2/Sit  this  method  is  entirely  satisfactory  unless  a 
is  more  than  10  percent  of  bd.  For  example,  let  the  width  b 
be  8  inches,  the  depth  d  be  12  inches,  and  each  of  the  four  steel 
rods  be  ij  inches  in  diameter,  the  centers  of  rods  being  4  inches 
from  the  neutral  axis;  then  bd  =  g6  square  inches  and  a  =4.91 
square  inches;  Ci=6.o  and  £2=4.62  inches,  71=1152  inches4, 
and  72=4.9iX42  =  79  inches4.  Now,  taking  E2/Ei  =  io,  the 
first  formula  of  (112)'  gives  82/81  =  about  8;  accordingly  if 
the  safe  allowable  tensile  unit-stress  for  concrete  is  taken  as 
50  pounds  per  square  inch,  ,$2  for  the  steel  will  be  only  400  pounds 
per  square  inch.  The  steel  and  concrete  do  not  work  well  together, 
and  in  fact  this  design  is  a  poor  one.  If  the  load  is  sufficiently 
large  to  make  £2  as  high  as  12  ooo  pounds  per  square  inch,  a 
value  which  the  steel  may  safely  bear,  the  concrete  will  be  rup- 
tured by  tension  on  the  convex  side,  so  that  it  can  only  serve 
as  a  kind  of  protective  covering  for  the  steel  rods.  The  uniform 
load  which  these  rods  can  safely  carry  will  then  be  Wz^&SJz/cJ. 
For  instance,  if  the  simple  beam  is  8  feet  4  inches  long  W%  =  16  400 
pounds  is  the  load  which  is  carried  by  the  steel  if  the  concrete 
does  not  act  at  all,  while  Wi=jjo  pounds  is  the  load  which 
might  be  safely  carried  by  the  concrete  without  any  reinforce- 
ment. Since  this  concrete  beam  weighs  about  800  pounds,  it 
would  safely  carry  only  its  own  weight  unless  reinforced. 

Figs.  113c  show  the  methods  of  reinforcement  generally  used 
for  concrete  beams,  the  metal  being  placed  only  on  the  tensile  side. 
The  theory  given  in  the  preceding  article  is  entirely  inapplica- 
ble to  a  beam  where  the  metal  is  only  on  one  side  of  the  neutral 
axis,  and  the  proper  theory  will  be  developed  in  the  two  following 
articles.  Various  forms  of  rods  are  in  use,  and  it  has  been  found 
that  smooth  rods  are  not  the  best,  since  there  is  a  tendency  for 
them  to  slip  in  the  concrete.  One  of  the  oldest  kinds  is  a  rect- 
angular twisted  bar  which  is  known  as  the  Ransome  rod ;  another 
form,  is  that  of  Thacher,  which  is  a  round  bar  flattened  in  two 
rectangular  directions;  Johnson's  bar  is  of  rectangular  section 
with  corrugations  alternating  on  adjacent  sides.  Another  rein- 
forcement is  the  lozenge-shaped  form,  called  expanded  metal, 


288  COMPOUND  COLUMNS  AND  BEAMS          CHAP.  XII 

which  is  widely  used  for  the  beams  of  floors.  The  Kahn  method 
consists  of  rods  which  are  straight  near  the  middle  of  the  beam 
and  bent  upward  near  the  ends,  the  inclined  ends  being  intended 
to  prevent  the  shearing  that  sometimes  occurs  along  surfaces 
which  are  indicated  by  the  shear  lines  in  Fig.  109. 

When  steel  rods  are  used  in  concrete  beams,  the  fundamental 
idea  is  that  they  are  for  the  purpose  of  increasing  the  resistance 
on  the  tensile  side.  A  plain  concrete  beam  has  its  neutral  sur- 
face at  the  middle  and  hence  the  compressive  unit-stress  on  the 
upper  surface  is  equal  to  the  tensile  unit-stress  on  the  lower 
surface  when  the  elastic  limit  is  not  exceeded.  Hence  in  a  plain 
concrete  beam  the  resistance  to  compression  cannot  be  developed 
until  the  beam  is  ruptured  on  the  tensile  side.  When  rods  are 
placed  near  the  lower  side,  these  take  tensile  stresses  and  hence 
the  compressive  stresses  in  the  concrete  above  the  neutral  sur- 
face are  increased.  In  this  case  the  neutral  axis  no  longer 
passes  through  the  center  of  gravity  of  the  section,  so  that  new 
formulas  must  be  established  for  concrete  beams  reinforced  in 
this  manner,  and  this  will  be  done  in  the  next  article. 

Concrete  beams  are  usually  rectangular  in  section,  and  only 
these  will  be  discussed  in  the  following  pages.  The  amount  of 
metal  which  is  placed  near  the  tensile  side  of  the  section  is  rarely 
greater  than  2  percent  of  the  total  rectangular  section  area,  about 
i  percent  being  the  usual  practice.  Structural  steel  is  mainly 
employed,  although  hard  steel  has  been  sometimes  used.  The 
allowable  unit-stress  for  structural  steel  is  generally  taken  as 
12  ooo  pounds  per  square  inch,  although  there  would  be  little 
objection  to  stressing  it  to  a  value  25  percent  higher,  but  it  is 
difficult  to  develop  the  full  resistance  of  the  steel,  as  will  be  seer, 
later.  In  the  discussions  of  the  following  pages  the  average 
values  of  the  moduluses  of  elasticity  given  in  Table  2  will  be 
used,  and  hence  the  ratio  E%/E\  will  be  taken  as  10.  The  allow- 
able working  unit-stresses  for  concrete  will  be  generally  taken 
as  500  pounds  per  square  inch  in  compression  and  60  in  tension; 
these  values  apply  only  to  first-class  concrete  of  the  proportions 
I  cement,  2  sand,  4  broken  stone  (Art.  22). 


ART.  114  THEORY  OF  REINFORCED  CONCRETE  BEAMS    289 

Prob.  113.  A  section  like  Fig.  1136  is  to  be  used  in  a  floor  for  a 
span  of  6J  feet,  the  depth  of  the  beam  being  5  inches  and  its  width 

4  feet.     The  steel  rods  are  one  inch  in  diameter  and  placed  so  that 
their  centers  are  ij  inches  from  the  neutral  axis.     How  many  rods 
are  needed  in  order  that  they  alone  may  carry  a  total  uniform  load  of 

5  850  pounds  with  a  factor  of  safety  of  5  ? 


ART.  114.     THEORY  OF  REINFORCED  CONCRETE  BEAMS 

The  theoretic  laws  of  Art.  39  apply  to  all  kinds  of  beams, 
as  also  the  experimental  laws  4  and  5  of  Art.  40.  When  different 
materials  are  used  in  a  beam  law  6  needs  modification,  for  the 
stiffest  material  is  stressed  the  highest.  Law  5  is  to  be  used 
in  this  case,  and  this  states  that  the  unit-elongation  or  unit- 
shortening  is  the  same  for  both  materials.  Thus  at  the  same 
distance  from  the  neutral  surface  S\/Ei  for  the  concrete  must 
tqual  5 2/^2  for  the  steel.  Law  6  applies,  however,  to  the  stresses 
in  each  material,  and  thus  the  stresses  in  the  concrete  vary  directly 
as  their  distances  from  the  neutral  surface.  The  same  is  true 
for  the  stresses  in  the  metal,  but  since  the  section  area  of  this 
is  small,  it  will  usually  be  sufficient  to  regard  the  unit-stresses 
in  the  metal  as  uniformly  distributed  (Art.  113).  Let  b  be  the 
width  and  d  the  depth  of  the  rectangular  section,  and  a  the  section 
area  of  the  metal.  In  strictness  the  section  area  of  the  concrete 
is  bd  —  a,  but  it  will  be  unnecessary  to  take  the  diminution  of  bd 
into  account,  since  a  is  rarely  greater  than  3  percent  of  bd. 

CASE  I.  Tension  in  Concrete.  Fig.  114#  shows  a  concrete 
beam  reinforced  with  horizontal  steel  bars  at  a  distance  h  below 
the  middle.  Let  Si  be  the  compressive  unit-stress  mp  on  the 
upper  side  at  the  dangerous  section,  S\  the  tensile  unit-stress 
m'p'  on  the  lower  side,  and  S2  the  tensile  unit-stress  in  the  steel. 
Let  ss  be  the  neutral  surface  at  the  distance  k  below  the  middle 
of  the  simple  beam.  The  first  condition  of  static  equilibrium 
is  that  the  algebraic  sum  of  all  the  horizontal  stresses  in  the  section 
shall  equal  zero;  the  sum  of  all  the  compressive  stresses  is 
for  the  mean  unit-stress  Si  acts  over  the  area 


290  COMPOUND  COLUMNS  AND  BEAMS  CHAP.  XII 


,  the  sum  of  all  the  tensile  stresses  in  the  concrete  is 
—  k),  the  sum  of  all  the  tensile  stresses  in  the  steel  i* 
The  second  condition  of  static  equilibrium  is  that  the 
sum  of  the  moments  of  these  stresses  shall  equal  the  bending 
moment  M  ;  the  moment  of  the  compressive  stresses  with  respect 
to  the  neutral  axis  ss'  is  %Sib(^d+k)2,  since  the  total  compressive 
stress  acts  with  the  lever  arm  §(Ji+£),  the  moment  of  the  tensile 
stresses  in  the  concrete  is  $Si'b(%d—k)2,  and  the  moment  of 
the  tensile  stresses  in  the  steel  is  S2a(h  —  k).  Accordingly, 


are  two  equations  connecting  the  four  unknown  quantities  Si, 
Si',  S2,  k.  Two  other  equations  result  from  the  law  that  the 
unit-elongations  are  proportional  to  their  distances  from  the 
neutral  surface:  Si-/Sif  =  Qd+k)/($d—K)  applies  to  the  concrete 
if  the  elastic  limit  is  not  exceeded;  S2/E2  is  the  unit-elongation 
in  the  steel,  and  this  equals  the  unit-elongation  in  the  concrete  at 
the  distance  h  —  k  from  the  neutral  surface,  hence  S2/E2  = 
Si(h  —  k)/Ei($d+k).  The  solution  of  the  four  equations  gives, 


_  _  _ 

(bd/a)(E,/E2)  b(d2+i2hk) 


014} 


from'  which  it  is  seen  that  the  position  of  the  neutral  surface 
depends  upon  h  and  the  ratios  a/bd  and  E2/!E.  When  h  =  o, 
the  steel  rod  is  at  the  middle  of  the  depth,  and  then  S2  =  o, 
Si=Si'  =  6M/bd2  which  are  the  unit-stresses  for  an  unreinforced 
beam.  The  steel  reinforcement  is  usually  placed  near  the  base. 

The  above  theory  applies  only  when  the  unit-stresses  do  not 
exceed  the  elastic  limit,  but  concrete  is  a  brittle  material  in 
which  the  elastic  limit  is  poorly  denned,  the  stress  diagram  being 
similar  to  that  of  cast  iron  or  brick.  However,  by  using  high 
factors  of  safety,  Si  and  Si'  may  be  made  sufficiently  low  so  that 
the  above  formulas  have  the  same  validity  as  the  common  flexure 
formula  has  when  applied  to  brittle  materials.  Owing  to  the 
Jow  tensile  strength  of  concrete,  beams  often  crack  on  the  tensile 


ART.  114 


THEORY  OF  CONCRETE-STEEL  BEAMS 


291 


side,  so  that  the  reinforcing  bars  carry  nearly  all  of  the  tensile 
stresses.  It  is  therefore  important  to  develop  the  formulas  which 
apply  to  such  a  case,  and  this  will  now  be  done  under  the  assump- 
tion that  all  the  tensile  stress  below  the  neutral  surface  is  carried 
by  the  steel  bars. 


m 

q 

™,,.  —  ,r 

4 

*  —  -77P 

A     j. 

t.     i"'     _           i! 

< 

rd     r 

! 

-rT 

3\     S2  ^ 

.... 

i 

t_ 

{                i 
i     ^               !     2  >- 

•                m/ 

y9 

-*-  6  > 

I 

a 

Fig.  114a  Fig.  1146 

CASE  II.  No  Tension  in  Concrete.  Let  Fig.  H4b  represent 
the  case  where  there  are  no  tensile  stresses  in  the  concrete,  the 
notation  being  the  same  as  before  except  that  the  position  of 
the  steel  is  designated  by  the  distance  g  measured  downward 
from  the  upper  surface  of  the  beam,  while  the  neutral  surface 
55  is  at  the  distance  n  below  the  same  surface.  The  compressive 
unit-stress  Si  on  the  upper  side  is  represented  by  mp,  while  S2 
is  the  tensile  unit-stress  in  the  metal.  To  determine  these  quan- 
tities, the  static  laws  of  Art.  39,  together  with  the  experimental 
laws  of  Art.  40,  are  again  to  be  used.  The  sum  of  all  the  horizontal 
stresses  below  the  neutral  axis  in  the  dangerous  section  is  S^a 
and  the  sum  of  those  above  it  is  ^S\bn,  since  the  average  unit- 
stress  JSi  acts  over  the  area  bn;  hence  S^a-=\S\})n  is  the  first 
equation  between  £2  and  Si.  The  sum  of  the  moments  of  all  the 
stresses  in  the  section  is  the  resisting  moment  which  equals  the 
bending  moment  M.  Now  Szd(g  —  n)  is  the  moment  of  the 
stresses  in  the  metal  with  respect  to  the  axis  s,  and  JSi&»2  is 
the  moment  of  the  stresses  in  the  concrete,  since  the  total  stress 
}>Sibn  acts  with  the  lever  arm  \n\  hence  S^(g  —  n)  +  \Sibn2  is 
the  resisting  moment  which  equals  M,  and  this  is  a  second  equa- 
tion between  S2  and  Si.  A  third  condition  is,  however,  required, 
since  the  unknown  quantity  n  is  contained  in  each  of  those  thus 
far  established.  This  is  furnished  by  the  experimental  law 
regarding  changes  of  length;  $2/E2  is  the  unit-elongation  of 
the  metal  which  is  at  the  distance  g— n  from  the  neutral  surface. 


292  COMPOUND  COLUMNS  AND  BEAMS  CHAP.  XII 

while  the  unit-shortening  of  the  concrete  at  the  same  distance 
from  the  neutral  surface  is  Si(g  —  n)/Ein.     Hence, 

S2a=  $Sibn        S2a(g-n}  +  $Sibn2=M        S2/E2=  (Si/Ei)(g-n)/n 
are  three  equations  for  finding  n,  S\,  S2]  their  solution  gives 


ea 

in  which  s  denotes  the  ratio  E2/Ei.  These  formulas  do  not 
contain  the  depth  d,  but  this  is  usually  i  or  i^  inches  greater 
than  g  in  order  to  protect  the  steel  from  corrosion. 

Prob.  114#.  A  reinforced  concrete  beam  is  5  inches  deep,  4  feet 
wide,  6J  feet  in  span,  has  1.8  square  inches  of  steel  at  ij  inches  belcw 
the  middle,  and  the  total  load  upon  it  is  6  400  pounds.  Show  from 
(114)  that  the  concrete  will  probably  crash  on  the  tensile  side. 

Prob.  1146.  Using  the  above  data  and  supposing  that  the  concrete 
offers  no  tensile  resistance,  compute  from  (114)'  the  position  of  the 
neutral  surface  and  the  unit-stresses  Si  and  S2. 

ART.  115.     INVESTIGATION  OF  REINFORCED  CONCRETE  BEAAIS 

The  formulas  of  the  last  article  furnish  the  means  of 
investigating  a  reinforced  concrete  beam  for  which  the  dimensions 
and  loads  are  given.  When  the  beam  is  lightly  loaded,  so  that 
the  concrete  below  the  neutral  surface  is  in  tension,  formulas 
(114)  are  to  be  used.  When  the  beam  is  so  heavily  loaded  that 
this  tensile  resistance  is  overcome,  formulas  (114)'  are  to  be 
used,  provided  the  elastic  limit  of  the  concrete  on  the  compressive 
side  is  not  exceeded.  This  elastic  limit  is  an  uncertain  quantity, 
but  it  is  probably  not  far  from  600  or  700  pounds  per  square 
inch  when  the  concrete  has  the  proportions  of  i  cement,  2  sand, 
4  broken  stone.  When  the  beam  is  so  heavily  loaded  that  the 
computed  Si  exceeds  this  elastic  limit,  the  formulas  do  not  give 
reliable  values  of  the  unit-  stresses. 

For  example,  let  a  reinforced  concrete  beam  be  5  inches  deep, 
4  feet  wide,  4^  feet  in  span,  and  have  3.6  square  inches  of  steel 
placed  2  inches  below  the  middle.  Let  it  be  required  to  investigate 


ART.  115        INVESTIGATION  OF  REINFORCED  BEAMS  293 

this  beam  when  it  carries  a  uniform  load  of  2  400  pounds,  includ^ 
ing  its  own  weight.  Here  £  =  48,  d  =  $,  h=2  inches;  6^=240 
and  0=3.6  square  inches,  whence  0/6^  =  0.015;  also  E2/Ei  =  io 
(Art.  113).  Using  formulas  (114)  there  is  found  £  =  0.261  inches 
for  the  location  of  the  neutral  surface  belov;  the  middle.  The 
maximum  moment  is  M  =  \Wl  =  i()  200  pound-inches,  vrhence 

5  =  6M/bd2  =  Si   pounds  per  square  inch  as  the  flexural  unit- 
stress  for  a  plain  concrete    beam.     Then  6*1  =  72,    and    5V  =  58 
pounds  per  square  inch  for  the  compressive  and  tensile  stresses 
on   the  upper  and  lower  surfaces  of  the  beam,  the  first  giving 
a  factor  of  safety  of  about  40,  while  the  second  gives  a  factor 
of  safety  of  about  5.     Also  £2  =  250  pounds  per  square  inch  for 
the  steel,  which  is  a  very  low  value,   the  factor  of  safety  being 
over  200.     While  this  beam  is  perfectly  safe,  it  is  not  designed 
for  proper  economy,  since  the  compressive  stress  in  the  concrete 
and  the  tensile  stress  in  the  steel  might  be  much  higher. 

As  another  example,  let  ^  =  48,  ^=5,  A=iJ,  g  =  4,  I  =  84 
inches,  a  =  2. 4  square  inches,  and  the  total  uniform  load  W  be 

6  ooo  pounds.     Taking  E2/Ei  =  10,    (114)  gives  Si'  =450  pounds 
per  square  inch,  which  is  greater  than  the  tensile  strength  of  the 
concrete,  so  that  these  formulas  do  not  apply.    Turning  then  to 
(114)'  there  are  found  n=  1.56  inches,  £1  =  373  pounds  per  square 
inch  for  the  concrete  and  52  =  1 1  700  pounds  per  square  inch 
for  the  steel,  so  that  the  beam  has  a  proper  degree  of  security. 

The  formulas  of  Art.  114  are  valid  when  the  unit-stresses  in 
the  concrete  are  proportional  to  their  distances  from  the  neutral 
surface,  and  this  is  the  case  only  when  the  changes  of  length 
are  proportional  to  the  stresses.  Concrete  is  a  material  in  which 
this  proportionality  exists  only  for  low  unit-stresses,  so  that  the 
validity  of  the  formulas  is  sometimes  questioned.  Hatt  has 
deduced  formulas  under  the  supposition  that  the  unit-stresses 
vary  with  their  distances  from  the  axis  according  to  a  parabolic 
law,  and  these  will  undoubtedly  give  a  better  agreement  with 
experiment  than  (114)'  when  the  concrete  is  highly  stressed. 
For  a  case  of  design,  however,  the  prevailing  opinion  is  that 
formulas  (116)  should  be  used,  and  this  has  been  the  common 


294  COMPOUND  COLUMNS  AND  BEAMS  CHAP.  XII 

practice  since  1900.  The  general  laws  involved  in  formulas  (114)' 
are  confirmed  by  experiments  in  which  beams  are  ruptured, 
although  numerical  values  computed  from  the  formulas  are  of 
little  reliability  except  as  empirical  results  similar  to  that  of  the 
computed  flexural  strength  or  modulus  of  rupture  for  beams  of 
one  material  (Art.  52). 

The  phenomena  of  failure  of  a  reinforced  concrete  beam  have 
been  completely  ascertained  by  the  very  valuable  experiments  made 
by  Talbot  in  1904.  These  beams  were  12  inches  wide,  13 \  inches 
deep,  14  feet  in  span,  and  had  the  steel  reinforcing  bars  12  inches 
below  the  top  surface.  Various  percentages  of  metal  were  used, 
ranging  from  0.42  to  1.56  percent  of  that  of  the  concrete,  and 
several  kinds  of  reinforcing  bars  were  employed.  The  beams 
were  tested  by  applying  two  concentrated  loads  at  the  third 
points  of  the  span,  and  the  deflections  at  the  middle  were  measured 
for  several  increments  of  loading,  as  also  horizontal  changes  of 
length.  Under  light  loads  the  tensile  resistance  of  the  concrete 
was  plainly  apparent;  when  the  tensile  unit-stress  in  the  concrete 
reached  about  350  pounds  per  square  inch,  the  neutral  surface 
rose  and  the  stress  in  the  steel  increased.  A  little  later  fine 
vertical  cracks  appeared  on  the  tensile  side,  while  the  tensile 
stresses  in  the  steel  and  the  compressive  stress  in  the  concrete 
increased  faster  than  the  increments  of  the  load.  The  last  stage 
was  a  rapid  increase  of  the  deformations,  and  rupture  occurred 
by  the  crushing  of  the  concrete  on  the  upper  surface,  the  steel 
being  then  generally  stressed  beyond  its  elastic  limit. 

In  some  cases  reinforced  concrete  beams  have  been  known  to 
fail  by  shearing  near  the  ends,  the  curve  of  rupture  being  like  that 
shown  by  the  broken  lines  in  Fig.  109.  The  full  investigation 
of  this  case  is  attended  with  some  difficulty  and  will  not  here  be 
attempted,  but  the  discussion  of  Art.  108  furnishes  the  means 
of  making  approximate  computations.  It  is  only  short  beams 
which  fail  in  this  manner.  When  rupture  occurs  near  the  middle 
of  the  beam  along  a  curved  surface  which  roughly  agrees  with 
one  of  the  full  lines  in  Fig.  109,  this  is  not  a  case  of  shearing 
but  one  of  rupture  by  tension. 


ART.  116      DESIGN  OF  REINFORCED  CONCRETE  BEAMS         295 

The  safe  load  which  may  be  carried  by  a  reinforced  concrete 
beam  can  be  computed  in  five  ways  from  the  formulas  of  Art. 
114.  Under  Case  I  the  first  step  is  to  find  k,  and  then  to  place 
its  value  in  the  three  following  equations.  Allowable  values  of 
Si,  Si',  and  £2  being  assumed,  three  computations  maybe  made  to 
find  three  values  of  S  from  which  three  values  of  M  are  determined; 
then  for  uniform  load  W=SM/l  and  the  smallest  of  the  three 
values  of  W  is  the  one  to  be  selected  on  the  theory  of  Case  I,  where 
no  cracking  of  the  concrete  is  allowed.  Larger  values  of  W  will 
be  found  by  using  the  formulas  of  Case  II,  the  first  step  being 
to  compute  n,  the  second  to  find  two  values  of  M  from  assumed 
values  of  Si  and  £2,  and  the  third  to  compute  two  values  of  W, 
the  smaller  of  which  is  the  safe  load.  Computations  generally 
show  that  a  reinforced  concrete  beam  is  stronger  after  it  has 
cracked  on  the  tensile  side  than  it  was  before,  this  being  due  to 
the  fact  that  the  steel  then  has  a  higher  unit-stress.  These  cracks 
may  not  be  visible  and  they  are  often  called  hair  cracks;  they 
exist  whenever  the  unit-stress  on  the  tensile  side  of  the  beam 
reaches  or  exceeds  the  ultimate  strength. 

Prob.  115a.  A  reinforced  concrete  beam  is  12  inches  wide,  15 
inches  deep,  14  feet  long,  and  has  3.6  square  inches  at  ij  inches  from 
the  lower  side.  Find  the  total  uniform  load  W  which  this  beam  can 
carry  so  that  the  tensile  stress  in  the  concrete  on  the  lower  side  may 
be  100  pounds  per  square  inch. 

Prob.  115&.  Using  the  same  data  as  above,  compute  the  total 
uniform  load  which  will  produce  a  compressive  stress  of  500  pounds 
per  square  inch  on  the  concrete,  considering  that  the  concrete  below 
the  neutral  surface  offers  no  tensile  resistance. 

ART.   116.     DESIGN  OF  REINFORCED  CONCRETE  BEAMS 

When  a  reinforced  concrete  beam  is  to  be  built,  its  width, 
depth,  and  span  are  given  or  assumed,  as  also  the  load  which 
is  to  be  carried  and  the  allowable  unit-stresses  for  the  concrete 
and  steel.  The  problem  of  design  then  consists  in  determining 
the  proper  section  area  of  the  reinforcing  bars  and  the  proper 
depth  of  the  beam.  As  for  the  position  of  these  bars,  it  is  ap- 
parent that  they  should  be  placed  as  near  as  possible  to  the  tensile 


296  COMPOUND  COLUMNS  AND  BEAMS  CHAP.  XII 

side  of  the  beam  in  order  that  their  resisting  moment  may  be  as 
large  as  possible.  They  should,  however,  be  entirely  covered 
by  the  concrete  in  order  to  be  protected  from  corrosion  due  t? 
atmospheric  influences. 

It  is  impossible  to  design  an  economical  reinforced  concrete 
beam  on  the  theory  of  Case  I  of  Art.  114,  for  if  Sf  be  taken  even 
as  high  as  the  ultimate  tensile  strength  of  the  concrete,  the  values 
of  Si  and  £2  are  too  low;  52  for  the  steel  is  indeed  always  less 
than  Si',  so  that  the  strength  of  the  metal  is  not  utilized.  Nothing 
remains  to  be  done,  therefore,  but  to  allow  the  concrete  to  crack 
on  the  tensile  side  and  thus  bring  proper  tension  into  the  steel. 
If  the  tensile  resistance  of  the  concrete  is  not  considered,  formulas 
(114)'  apply.  The  given  quantities  are  E%/Ei  or  e,  the  width  b, 
the  unit-stresses  Si  and  Si,  and  the  bending  moment  M.  Eliminat- 
ing n  from  the  three  equations,  there  are  found  two  equations 
containing  g  and  a,  and  the  solution  of  these  gives 


in  which  a  denotes  the  ratio  S2/Si.  The  unit-stress  Si  should 
be  taken  as  high  as  allowable  ;  S2  should  not  be  higher  than 
the  allowable  value,  but  it  may  be  taken  lower,  if  economy  in 
cost  is  thereby  promoted.  The  depth  d  is  made  about  ij  inches 
greater  than  g. 

For  example  a  beam  is  to  be  built  of  concrete  which  has  the 
proportions  i  cement,  3  sand,  6  stone,  for  which  the  ratio 
E2/Ei  =  e=  15.  The  span  is  to  be  14  feet,  the  breadth  20  inches, 
and  the  total  uniform  load  is  to  be  7  ooo  pounds.  It  is  required 
to  find  the  depth  of  the  beam  and  the  section  area  of  the  reinforc- 
ing steel  rods  so  that  the  unit-stresses  Si  and  S2  shall  be  350  and 
14000  pounds  per  square  inch  respectively.  Here  #=40,  and 
M  =JX  7  000X14X12=  147  ooo  pound-inches.  Inserting  the 
given  values  in  the  fires  of  the  above  formulas,  there  is  found 
£=13.0  inches,  then  bg=26o  square  inches,  and  from  the  second 
formula  a  =  0.90  square  inches.  Here  the  section  area  of  the 
steel  is  0.35  percent  of  the.  section  of  the  beam  above  the  centers 


ART.  116     DESIGN  OF  REINFORCED  CONCRETE  BEAMS  297 

of  the  rods.  By  using  a  lower  value  of  52,  the  depth  g  will  be 
smaller  and  the  section  area  a  will  be  greater  than  the  above 
values.,  The  best  set  of  values  will  be  those  which  render  the 
cost  of  the  beam  a  minimum. 

The  position  of  the  neutral  surface  depends  only  upon  the 
values  of  the  ratios  e  and  a.  The  value  of  n,  as  found  from  the 
solution  of  the  equations  (114)',  is 

or  n=—g  (116)' 

- 


i+a/e- 

For  instance,  in  the  case  of  the  last  paragraph,  where  ^  =  15  and 
(7  =  40,  the  value  of  n  is  0.385  g,  so  that  the  neutral  surface  is  5.0 
inches  below  the  upper  side  of  the  beam. 

Steel  is  the  only  material  which  has  been  advantageously 
used  for  reinforcing  rods  in  concrete  beams,  and  the  proper 
section  area  should  rarely  exceed  one  percent  of  that  part  of  the 
concrete  above  the  centers  of  the  rods.  With  unit-stresses  of 
5  ooo  and  500  pounds  per  square  inch  for  steel  and  concrete 
respectively,  the  section  area  of  the  steel  is  2^  percent  of  that 
of  the  concrete  when  the  ratio  s  is  10  and  3  percent  when  £  is  15, 
but  this  is  an  excessive  use  of  steel  which  is  not  economical.  The 
following  are  computed  values  of  a/bg  in  percentages,  and  also 
values  of  the  ratio  n/g: 

S2/Si  =       15  20  25  30  35 

a/bg=     1.33  0.83  0.58  0.42  0.32 

n/g=     0.40  O-33  0.29  0-25  0.22 

These  apply  to  first-class  concrete,  for  which  e=io.  As  a  general 
rule  the  most  economic  dimensions  will  be  obtained  by  taking  S2 
as  about  30  times  Si  for  concrete  made  of  i  cement,  2  sand, 
3  stone,  while  S2  should  be  taken  as  about  12  500  and  Si  as 
about  350  pounds  per  square  inch  for  concrete  of  i  cement, 
3  sand,  6  stone,  the  ratio  S2/Si  being  here  35. 

Beams  having  a  section  like  that  of  Fig.  113d  are  sometimes 
used,  it  being  considered  that  only  the  upper  part  of  the  concrete 
is  stressed  in  compression,  while  the  tensile  stresses  in  the  lower 
projecting  parts  are  neglected.  The  design  of  these  is  hence 
made  by  the  formulas  (114)',  where  n  is  the  depth  of  the  upper 


298  COMPOUND  COLUMNS  AND  BEAMS  CHAP,  xn 

slab,  g  the  depth  from  its  upper  surface  to  the  center  of  the  steel 
rods,  and  b  the  width  of  slab  belonging  to  one  rod.  These 
formulas  may  be  written 

/      (j  \  i  bn 

£~n(I     £)  ~~2~a 

from  the  first  of  which  g  is  to  be  found  for  a  given  depth  n, 
while  the  second  gives  the  metal  section  a.  For  example,  let 
E2/Ei  =  s  =  io,  S2/Si=o  =  2$,  n  =  4  and  b  =  iS  inches;  then 
£=14.0  inches  and  (3=1.42  square  inches.  The  safe  load  which 
this  beam  can  carry  may  be  found  from  M=S^a  g  —  Jn);  for 
instance,  in  the  above  case  let  S2=  12  500  pounds  per  square 
inch,  then  the  safe  bending  moment  is  22  ooo  pound-inches. 

Prob.  116.  Consult  a  paper  by  Sewall  and  the  accompanying 
discussions  in  Transactions  of  American  Society  of  Civil  Engineers, 
1906,  Vol.  56.  Ascertain  different  opinions  as  to  what  should  be  the 
comparative  cost  of  concrete  and  steel  in  order  to  produce  the  most 
economical  reinforced  concrete  beam. 

ART.  117.     PLATE  GIRDERS 

A  plate  girder  is  composed  of  only  one  material,  usually 
structural  steel,  but  it  may  be  called  compound  in  the  sense 
that  it  consists  of  different  parts  riveted  together.  Fig.  117 
shows  a  side-view  and  section  of  a  plate  girder  without  the  rivets 
which  connect  the  web  to  the  angles  and  the  angles  to  the  cover 
plates.  In  the  section  A  a  there  are  four  angles  and  the  web; 
to  the  right  of  the  section  Bb  there  are  in  addition  two  cover 
plates;  to  the  right  of  the  section  Cc  there  are  four  cover  plates 
in  addition  to  the  angles  and  web.  The  section  areas  and 
moments  of  inertia  are  hence  different  in  the  three  sections,  the 
plate  girder  being  in  fact  an  approximation  to  a  beam  of  uniform 
strength  (Art.  58). 

The  flexure  formula  (41)  may  be  used  to  investigate  the 
strength  of  a  plate  girder  in  exactly  the  same  manner  as  if  it 
were  a  solid  beam.  Let  /i  be  the  moment  of  inertia  of  the  uni- 
form section  between  A  a  and  Bb,  I2  that  between  Bb  and  Cc, 
and  /s  that  between  Cc  and  Dd.  Then  S=  MI  .  Ci/Ii  applies 
to  the  first  section,  S=M2.c2/I2  to  the  second  section,  and 


ART  117  PLATE  GIRDERS  299 

so  on.  By  the  help  of  this  formula  sections  may  also  be  found 
to  resist  the  bending  moments  under  a  specified  unit-stress  S. 
For  example,  let  the  load  be  uniform  and  expressed  by  w  per 
linear  unit;  let  /  be  the  length  of  the  beam,  and  the  three  dis- 
tances ab,  ac,  ad,  be  o.2/,  0.35^  and  o-5/.  Then  from  Art.  38, 
the  bending  moment  at  Bb  is  o.o&owl2,  that  at  Cc  is  o.n^.wl2, 
and  that  at  Dd  is  0.12 $wl2.  Hence  the  value  of  I\/c\  is  o.o&wl2/S, 
that  of  /2/C2  is  o.ii4.wl2/S,  and  that  of  1 3/£3  is  o.i2$ivl2/S. 
Sections  of  plates,  angles  and  web  may  now  be  determined,  with 
the  help  of  Art.  44  and  Table  10,  to  satisfy  these  requirements. 
Then  the  section  Aa  must  be  investigated  to  ascertain  whether 
it  is  Fufficiertly  large  to  safely  resist  the  vertical  shear  \wl. 


Fig.  117 

Another  method  which  is  frequently  used  in  practice  is  to 
regard  the  web  as  carrying  none  of  the  bending  moment,  and  to 
consider  that  the  unit-stresses  in  the  flanges  are  uniformly  dis- 
tributed so  that  the  total  stress  in  each  flange  may  be  regarded 
as  acting  at  its  center  of  gravity.  Let  d  be  the  depth  between 
the  centers  of  gravity  of  the  flanges,  a  the  section  area  of  one 
flange,  and  S  the  allowable  unit-stress.  Then  the  stress  Sa 
in  one  flange  acts  with  the  lever  arm  d  with  respect  to  the  center 
of  gravity  of  the  other  flange,  and  therefore  Sad  equals  the  bend- 
ing moment  M.  Thus,  for  the  first  section  ai=M\/Sd\,  for 
the  second  section  a2=M2/Sd2,  and  so  on;  from  these  values 
of  a  proper  angles  and  plates  may  be  selected.  The  section 
area  of  the  web  is  determined  in  this  method  from  the  maxi- 
mum vertical  shear  which  can  act  at  the  end.  The  thickness 
of  the  web  is  made  uniform  throughout  the  span;  f,  \  and  f 
inches  are  common  thicknesses,  as  these  are  standard  market 
sizes.  The  web  is  usually  stiffened  by  vertical  angles  riveted  to 
it  at  intervals. 


300  COMPOUND  COLUMNS  AND  BEAMS  CHAP.  XII 

After  a  thickness  has  been  determined  for  the  web  from  the 
vertical  shear,  it  cannot  generally  be  altered  if  the  depth  of  the 
girder  is  slightly  changed  on  account  of  the  requirement  that 
market  sizes  shall  be  used.  There  is  then  a  certain  depth,  called 
the  "  economic  depth,"  which  gives  a  smaller  amount  of  mate- 
rial than  any  other.  For  the  simple  case  where  the  flange  areas 
are  uniform  throughout,  there  being  no  cover  plates,  this  eco- 
nomic depth  may  be  determined  in  the  following  manner.  The 
section  area  of  each  flange  is  M/Sd,  and  the  section  area  of  the 
web  is  td  The  total  volume  of  material,  neglecting  rivets,  splices, 
and  stiffeners.  then  is  (2M/Sd  +  td)l.  Differentiating  this  with 
respect  to  d,  and  equating  the  derivative  to  zero,  gives  d2  =  2M/St, 
which  determines  the  economic  depth.  This  condition  shows 
that  2M/Sd  equals  td,  that  is,  the  girder  has  its  economic  depth 
when  the  amount  of  material  in  the  flanges  is  equal  to  that  in 
the  web.  This  rule  holds  approximately  when  cover  plates 
are  used,  as  shown  by  the  investigations  in  Part  III  of  Roofs 
and  Bridges,  where  are  also  given  in  full  detail  the  methods 
of  designing  plate  girders  for  stringers,  floor  beams,  and 
bridges. 

Prob.  117.  A  plate  girder  used  as  a  floor  stringer  has  a  span  of  22 
feet,  and  the  uniform  load  w  per  linear  foot  which  is  equivalent  to  the 
actual  wheel  loads  is  i  700  pounds.  For  an  effective  depth  d  of  34 
inches,  compute  the  flange  areas  at  the  middle  and  the  quarter  sec- 
tions, taking  S  as  1 2  ooo  pounds  per  square  inch. 


ART.  118.     DEFLECTION  OF  COMPOUND  BEAMS 

The  deflection  of  flitched  beams,  like  those  of  Art.  112,  is 
readily  computed,. when  the  elastic  limit  of  the  material  is  not 
exceeded,  by  the  use  of  the  formula  f=Wil3/aEi!i  in  which 
Wi  is  the  total  load  that  comes  on  the  material  that  has  the 
modulus  of  elasticity  EI  and  the  moment  of  inertia  /i.  The 
same  method  applies  to  the  compound  beams  of  Fig.  113a  and 
1136,  but  it  does  not  apply  when  reinforcing  bars  are  placed 
only  on  one  side  of  the  neutral  surface  of  a  concrete  beam.  Con- 


ART.  118  DEFLECTION  OF  COMPOUND  BEAMS  301 

crete-steel  beams  are  usually  of  short  span,  and  it  is  rarely  neces- 
sary to  compute  deflections.  Formulas  might,  indeed,  be  devised 
for  this  case,  but  they  would  be  of  uncertain  application  on 
account  of  the  uncertainty  in  the  value  of  EI  and  because  resist- 
ing tensile  stresses  might  not  exist  near  the  middle  of  the  beam, 
while  they  would  act  near  the  ends. 

For  plate  girders  it  is  sometimes  important  to  compute  the 
deflection  at  the  end  of  a  cantilever  arm  or  at  the  middle  of  a 
simple  span.  For  a  simple  span  under  uniform  load  the  deflec- 
tion found  in  Art.  55  is  / '=  $wl4 / ^S^EI ',  which  applies  to  a  plate 
girder  where  the  moment  of  inertia  /  is  constant  throughout. 
If  /i  is  the  moment  of  inertia  of  the  section  near  the  end  and 
/3  that  of  the  section  at  the  middle,  as  in  Fig.  117,  then  for  these 
values  there  may  be  found  two  deflections  f\  and  /2,  the  first  of 
which  is  greater  and  the  second  less  than  the  true  deflection.  It 
is  often  the  case  that  this  information  is  all  that  is  required,  but 
by  the  method  of  Art.  124  a  formula  giving  a  closer  result  can 
be  deduced.  Let  /i,  12,  Is  be  the  distances  from  the  left  end 
in  Fig.  117  to  the  sections  Bb,  Cc,  and  Dd,  so  that  /*3  is  one-half 
of  the  span  /.  Let  the  load  be  uniform  so  that  the  bending 
moment  at  the  distance  x  from  the  support  is  M  =  \wlx  —  %wx2. 
Let  m  be  the  bending  moment  due  to  a  load  unity  at  the  middle 
of  the  beam  or  m  =  \x.  Then,  by  (124), 

,       fMm.  rtiMm-  i^Mm.  f^Mm. 

f=     I    ~=^3X=2    I       —§X+2    I        _J#+2/       -^ffa 

J     EI  J0     Eli  Jk    EI2  Ji2    EI3 

gives  the  deflection  at  the  middle  under  the  uniform  load.  Inte- 
grating between  the  designated  limits,  there  is  found, 

f=wl    //I3      /23~/l3  .   /33~/23\          W    //!4      /24-*l4,/34-/2*\ 

J     6E\/i         /2  73     /      8EWi         /2  /s    / 

This  formula  is  not  difficult  in  computations  when  tables  of 
squares,  cubes,  and  reciprocals  are  at  hand ;  thus  /i3//i  is  found 
by  taking  /i3  from  the  table  and  multiplying  it  by  the  reciprocal 
of /i. 

The  above  gives  the  elastic  deflection  due  to  the  horizontal 
flexural  stresses  only.  Art.  125  shows,  however,  that  there  is 
a  deflection  due  to  the  vertical  shears  which  must  be  added  to 


302  COMPOUND  COLUMNS  AND  BEAMS          CHAP.  Xll 

the  above  in  order  to  obtain  the  exact  deflection.  Let  V  be  the 
shear  due  to  the  given  uniform  load  and  v  be  the  shear  due 
to  the  load  unity  at  the  middle  of  the  beam,  so  that  V  =  \wl—  wx 
and  v  =  \.  Then,  by  (125), 


f  rvv*     rliVv  *     chvv  *     rhvv  * 

/=   /  -=-dx  =  2  I     -=r-dx+2  I     ^r-dx+2   I     r=—dx 
J   Fa  J0    F(n  Jk   Fa2          Jh   Fa3 

in  which  #1,  a2,  a3  are  the  section  areas  whose  moments  of  inertia 
are  /i,  /2,  /3,  and  F  is  the  shearing  modulus  of  elasticity.  Per- 
forming the  integrations,  there  is  found, 


, 


2F  \ai  d2  ^3     /         2F 

which  is  the  deflection  at  the  middle  of  the  beam  under  the  uni- 
form load  due  to  the  vertical  shears.  The  numerical  value  of 
this  shearing  deflection  is  usually  small  compared  to  that  due 
to  the  bending  moments,  but  in  short  spans  it  is  an  appreciable 
quantity. 

Prob.  118.  Deduce,  from  the  above  formulas,  the  deflections  due 
to  vertical  shears  and  bending  moments  when  the  simple  beam  has  the 
constant  section  area  a  and  the  constant  moment  of  inertia  /.  Show 
that  the  length  of  beam  for  which  these  two  deflections  are  equal  is 
given  by  (l/r)2  =  4&E/$F,  where  r  is  the  radius  of  gyration  of  the 
section. 


ART.  119      EXTERNAL  WORK  AND  INTERNAL  ENERGY  303 

/ 
CHAPTER  XIII 

RESILIENCE  AND  WORK 

ART.  119.     EXTERNAL  WORK  AND  INTERNAL  ENERGY 

When  a  force  is  applied  to  a  body  it  overcomes  a  resistance 
through  a  certain  distance  and  thus  external  work  is  performed 
on  the  body.  It  is  usually  the  case  that  the  force  is  applied  by 
increments,  so  that  it  increases  uniformly  and  gradually  from 
o  up  to  its  full  value  P.  When  a  tensile  load  is  applied  in  this 
manner  to  a  bar  producing  the  elongation  e,  the  work  performed 
is  \Pe,  or  equal  to  the  mean  load  \P  multiplied  by  the  distance  e. 
This  is  otherwise  seen  from  Fig.  14a,  where  the  shaded  area 
represents  the  work  performed  while  the  load  increases  from 
o  to  P.  Similarly,  in  the  case  of  a  beam  under  a  single  load, 
the  load  increases  from  o  to  P  and  produces  the  deflection  /t 
so  that  the  work  performed  is  \Pf.  This  work  done  upon  the 
bar  or  beam  is  called  the  '  external  work '. 

When  the  elasticity  of  the  body  is  not  impaired  and  the  load 
is  applied  so  gradually  that  no  work  is  expended  in  producing 
heat,  there  is  stored  within  the  body  an  amount  of  energy  equal 
to  the  external  work.  This  is  called  '  internal  energy  ',  or  some- 
times 'internal  potential  energy',  because  this  energy  may  be 
utilized  to  perform  an  amount  of  work  equal  to  the  external 
work  performed  upon  it.  When  the  elastic  limit  of  the  material 
is  not  exceeded,  the  internal  energy  in  a  stressed  bar  is  equal 
to  \Pe  and  that  in  a  beam  stressed  by  a  single  load  is  equal  to 
\Pf.  That  these  statements  are  correct,  many  experiments 
can  testify,  and  they  also  follow  from  the  law  of  conservation  of 
energy. 

The  internal  energy  which  can  be  stored  in  a  metal  bar  is 
very  small  compared  with  that  which  is  stored  in  a  mass  of  steam 
or  compressed  air  of  the  same  weight.  For  example,  take  a 
bar  of  structural  steel,  6  square  inches  in  section  area  and  25  feet 


304  RESILIENCE  AND  WORK  CHAP,  xill 

long.  The  load  P  which  will  stress  this  bar  to  its  elastic  limit 
is  P=6X35  000  =  210  ooo  pounds,  and  the  elongation  under  this 
load  is  e  =  35  000X25/30000000=0.0292  feet;  hence  the  work 
stored  in  the  bar  is  K  =  %X 210  000X0.0292  =3060  foot-pounds. 
This  bar  weighs  8^X6X10X1.02=510  pounds  (Art.  17),  and  an 
equal  weight  of  air  will  occupy  about  6  320  cubic  feet.  If  this  air 
is  in  a  pipe  of  one  foot  diameter,  it  will  fill  a  length  of  8  040  feet, 
and  under  a  pressure  of  15  pounds  per  square  inch  above  the 
atmospheric  pressure,  it  may  be  compressed  to  half  this  length. 
The  force  P  here  is  i  700  pounds,  and  the  energy  stored  in  the 
air  is  JXi  700X4020  =  3  517  ooo  foot-pounds,  which  is  nearly 
i  200  times  as  great  as  that  stored  in  the  steel  bar.  The  mate- 
rials of  construction  cannot,  therefore,  be  advantageously  used 
for  the  storage  of  energy. 

If  the  above  steel  bar  is  used  as  a  simple  beam  with  a  single 
load  at  the  middle,  the  section  being  square,  the  load  P  which 
will  stress  it  to  the  clastic  limit  is  found  from  the  flexure  formula 
(41)  to  be  P  =  i  143  pounds.  The  deflection  /  under  this  load 
is  found  from  Art.  55  to  be  7=7.14  inches.  Hence  the  energy 
stored  in  the  beam  is  JXi  143X7.14/12=340  foot-pounds,  or 
only  one-ninth  of  that  stored  in  the  bar.  Springs  in  the  form 
of  beams  are  used  on  vehicles  to  lessen  the  shocks  which  occur 
during  motion,  but  they  cannot  be  used  to  store  energy  for  the 
propulsion  of  a  vehicle,  on  account  of  the  great  weight  which 
would  be  required. 

When  a  bar  is  already  under  a  load  PI  which  has  produced 
the  elongation  e\>  an  additional  load  P2  produces  the  elongation 
e2)  so  that  the  total  energy  stored  in  the  bar  is  J(Pi  +  P2)  (ei  +  e2), 
as  clearly  appears  from  Fig.  1195.  When  P2  is  removed,  the 
work  that  can  be  performed  by  the  stored  energy  is  that  repre- 
sented by  the  shaded  area  or  %(2Pi+P2)e2.  Or,  if  the  load  on 
a  bar  ranges  from  the  lower  value  PI  to  the  higher  value  P,  as 
in  Fig.  119c,  the  elongation  being  e\  for  the  former  and  e  for 
the  latter,  then  the  work  K  performed  on  the  bar,  or  the  energy  K 
stored  in  it  by  the  increase  of  PI  to  P,  is  given  by  K  =  \Pe  — \P\e\. 
Similarly,  when  a  beam  is  under  a  single  load  which  increases 


ART.  119      EXTERNAL  WORK  AND  INTERNAL  ENERGY 


30o 


from  PI  to  P,  while  the  deflection  increases  from  fi  to  /,  the 
external  work  and  the  stored  energy  due  to  this  increase  are 
both  expressed  by  K  =  %Pf-%P1f1. 

When  a  part  of  a  bar  is  considered  which  has  a  length  of 
unity  and  a  section  area  of  unity,  the  load  P  is  the  unit-stress  S 
and  the  elongation  e  is  the  unit-elongation  e.  Thus, 

K=$Se  K=$Se-lSiei  (119) 

are  the  expressions  for  the  work  performed  on  one  cubic  unit 
of  the  material,  the  first  being  for  the  case  where  the  unit -stress 
increases  from  o  up  to  S  and  the  second  for  the  case  where  the 
unit-stress  increases  from  Si  up  to  S. 


Fig.  119a 


Fig.  1196 


Fig.  119c 


When  the  bar  is  stressed  so  that  S  exceeds  the  elastic  limit 
of  the  material,  the  elongations  increase  more  rapidly  than  the 
stresses  and  the  above  formulas  are  inapplicable.  Art.  14  shows 
that  the  external  work  required  for  rupture  is  very  large  com- 
pared with  that  required  to  stress  the  bar  up  to  its  elastic  limit, 
particularly  for  wrought  iron  and  steel.  The  area  in  Fig.  I4c 
between  the  curve  and  the  axis  of  elongations  measures  this 
external  work.  For  wrought  iron  and  steel  the  portion  of  the 
area  below  the  elastic  limit  is  so  small  that  it  may  be  disregarded, 
and  then  the  area  may  be  roughly  expressed  as  that  of  a  trape- 
zoid  having  the  length  e  equal  to  the  ultimate  elongation,  and 
limited  by  the  two  ordinates  which  represent  the  elastic  limit 
and  the  ultimate  strength.  Taking  the  elastic  limit  as  one-half 
of  the  ultimate  strength,  the  area  of  this  trapezoid  is  K  =  %sSu, 
which  is  the  work  required  to  rupture  by  tension  one  cubic  unit 
of  the  bar.  For  example,  take  the  two  specimens  of  unannealed 
and  annealed  Bessemer  steel  in  the  table  of  Art.  25,  which  had 
5^  =  125000  and  £=o.n  before  annealing  and  Su  =  99000  and 


306  RESILIENCE  AND  WORK  CHAP,  xni 

6=0.19  after  annealing;  here  #  =  10300  inch-pounds  for  the 
first  and  K  =  i4  100  inch-pounds  for  the  second,  so  that  the 
process  of  annealing  increased  37  percent  the  capacity  of  the 
steel  to  withstand  external  work.  Another  formula  sometimes 
used  for  unit  rupture  work  is  K  =%£(Se  +  2Su),  where  Se  is  the 
Clastic  limit  and  Su  the  ultimate  strength. 

Prob.  1190.  How  many  foot-pounds  of  work  are  required  to  stress 
a  wrought-iron  bar,  4  inches  in  diameter  and  54  inches  long,  from 
6  ooo  up  to  12  ooo  pounds  per  square  inch  ? 

Prob.  1196.  If  this  bar  is  used  as  a  beam  with  a  load  at  the  middle, 
how  many  foot-pounds  of  work  are  required 'to  increase  the  greatest 
unit-stress  at  the  dangerous  section  from  6  ooo  up  to  12  ooo  pounds 
per  square  inch? 

ART.  120.     RESILIENCE  OF  BARS 

The  term  "  Resilience  "  is  frequently  used  to  designate  the 
work  that  can  be  obtained  from  a  body  under  stress  when  it  is 
relieved  of  its  load.  When  the  elastic  limit  of  the  material  is 
not  exceeded  this  work  must  be  that  stored  within  the  bar  in 
the  form  of  stress  energy.  In  Art.  14,  as  well  as  in  Art.  119, 
it  was  shown  that  the  external  work  performed  in  elongating 
or  shortening  a  bar  is  %Pe,  and  this  is  the  resilience  which  may 
be  utiUzed  when  the  bar  is  entirely  relieved  from  stress.  Let 
the  section  area  of  the  bar  be  a  and  the  uniform  unit-stress  be  5, 
then  P  =  aS ;  also  let  the  length  of  the  bar  be  /  and  the  modulus 
of  elasticity  of  the  material  be  E,  then  the  change  of  length  is 
e  =  (S/E)L  Hence,  the  elastic  resilience  of  the  bar  is, 

K=\Pe  or  K=%(S2/E)al  (120) 

and  the  factor  %S2/E  is  called  the   'modulus  of  resilience  '  of 
the  material  when  5  is  the  unit-stress  at  the  elastic  limit. 

The  following  are  average  values  of  the  modulus  of  resilience 
of  materials  which  have  been  computed  from  the  average  con- 
stants given  in  Arts.  4  and  9: 

for  timber,  %S2/E  =  3.0  inch-pounds  per  cubic  inch 

for  cast  iron,         $S2/E  =   1.2  inch-pounds  per  cubic  inch 


ART.  120  RESILIENCE  OF  BARS  30? 

for  wrought  iron,  %Se2/E  =  i2.$  inch-pounds  per  cubic  inch 

for  structural  steel,         %S2/E  =  20.4  inch-pounds  per  cubic  inch 

Resilience  is  a  measure  of  the  capacity  of  a  body  to  resist  external 
work,  and  the  higher  the  modulus  of  resilience  the  greater  is 
the  capacity  of  a  material  both  to  store  up  energy  and  to  resist 
work  that  may  be  performed  upon  it.  The  modulus  of  resilience 
measures  this  capacity  up  to  the  elastic  limit  only.  The  total 
elastic  resilience  of  a  bar  is  found  by  multiplying  the  modulus 
of  resilience  by  the  volume  of  the  bar,  as  (120)  shows.  Thus, 
a  bar  of  structural  steel,  6  square  inches  in  section  area  and 
25  feet  long,  has  a  volume  of  i  800  cubic  inches,  and  hence  its 
elastic  resilience  is  36  720  inch-pounds  or  3  060  foot-pounds. 

When  a  bar  is  stressed  by  a  load  which  increases  from  PI 
to  P,  the  unit-stress  increases  from  Si  to  S,  and  by  (120)  the 
resilience  of  the  bar  when  the  load  is  decreased  from  P  to  PI  is 

K  =  l(S2/E)al-l(Si2/E)al=l(S2-Si2)/E  .  al 

Here,  as  before,  the  unit-stress  S  must  not  be  greater  than  the 
elastic  limit  of  the  material  of  the  bar. 

The  word  resilience  implies  a  spring,  and  it  should  not  be 
used  except  for  that  part  of  the  applied  work  which  can  be  recovered 
when  the  load  is  removed.  When  the  elastic  limit  is  exceeded 
and  the  load  is  released,  the  expression  %(S2/E)al  also  applies, 
as  shown  in  Art.  14,  to  the  work  that  can  be  utilized,  but  numerical 
values  of  this  resilience  are  of  no  importance  when  S  is  the  ulti- 
mate strength,  because  the  capacity  of  a  material  to  withstand 
external  work  is  properly  measured  by  the  product  of  its  ulti- 
mate strength  and  ultimate  elongation,  as  explained  at  the  close 
of  Art.  119. 

Prob.  120<z.  What  horse-power  engine  is  required  to  stress,  250 
times  per  minute,  a  bar  of  wrought  iron  18  feet  long  and  2  inches  in 
diameter  from  o  up  to  its  elastic  limit  ? 

Prob.  1206.  What  horse-power  engine  is  required  to  stress,  250 
times  per  minute,  a  bar  of  wrought  iron  18  feet  long  and  2  inches  in 
diameter  from  12  500  up  to  25  ooo  pounds  per  square  inch? 


308  RESILIENCE  AND  WORK  CHAP.  XIII 


ART.  121.    RESILIENCE  OF  BEAMS 

When  a  cantilever  beam  deflects  under  the  action  of  a  load  at 
the  end  or  a  simple  beam  deflects  under  a  load  at  the  middle, 
the  external  work  is  $Wf  as  long  as  the  deflection  /  increases 
proportionally  to  the  load  which  is  applied  by  increments  so 
that  it  increases  gradually  from  o  up  to  the  value  W  (Art.  119). 
The  resilience  of  the  beam  then  equals  %Wf  and  an  expression 
for  its  value  in  terms  of  the  flexural  unit-stress  S  may  be  obtained 
by  substituting  for  W  and  /  their  values  from  Art.  56.  Let  / 
be  the  length  of  the  beam,  c  the  distance  from  the  neutral  sur- 
face to  the  upper  or  lower  side  of  the  beam  where  the  unit-stress 
is  S,  and  /  the  moment  of  inertia  of  the  cross-section;  then, 

W=aSI/cl  f=aSl2/pcE 

where  a  is  i  for  a  cantilever  loaded  at  the  end  and  4  for  a  simple 
beam  loaded  at  the  middle,  while  /?  is  3  for  the  cantilever  and  48 
for  the  simple  beam.  Replacing  /  by  ar2,  where  a  is  the  section 
area  and  r  the  radius  of  gyration  of  that  section  with  respect  to 
the  neutral  axis,  the  elastic  resilience  of  the  beam  is, 

K=^Wf=(a2/p)(r/c)2%(S2/E)  .  al 

in  which  %(S2/E)  is  the  modulus  of  resilience  of  the  material 
and  al  is  the  volume  of  the  beam. 

For  either  a  cantilever  or  a  simple  beam  the  value  of  a2//? 
is  J.  For  a  rectangular  section  the  value  of  (r/c)2  is  J;  hence 
for  a  rectangular  beam  under  a  single  load,  the  elastic  resilience 
is  K=%-^(S2/E)al,  which  is  one-ninth  of  that  of  a  bar  under 
tension  or  compression.  For  a  circular  section,  the  value  of 
(r/c)2  is  J,  and  it  hence  follows  that  its  elastic  resilience  under 
a  single  load  is  one-twelfth  of  that  of  a  bar  under  axial  stress. 
For  the  /  sections  in  Table  6  the  value  of  (r/c)2  is  about  T\,  so 
that  their  resilience  is  greater  than  that  of  rectangular  or  cir- 
cular sections. 

When  a  beam  is  uniformly  loaded  with  w  per  linear  unit  the 
load  on  any  short  length  dx  is  w  .  dx,  and  if  y  is  the  deflection 
at  the  point  whose  abscissa  is  x,  the  elementary  external  work 
for  a  gradually  applied  load  is  %wy  .  dx.  The  integration  of 


121  RESILIENCE  OF  BEAMS  309 

this  over  the  entire  length  of  the  beam  will  give  the  total  external 
work  of  the  uniform  load.  For  example,  take  a  cantilever  loaded 
uniformly;  the  value  of  y  is  given  by  the  equation  of  the  elastic 
curve  in  Art.  54,  and  accordingly, 

K  =  ( 

is  the  external  work  of  the  uniform  load.  Substituting  for  W 
its  value  in  terms  of  S,  and  for  /  its  value  ar2,  the  resilience  of  a 
rectangular  cantilever  under  uniform  load  is  found  to  be  K  = 
j|.|(52/£)a/,  which  is  three-fifths  of  that  found  for  the  concen- 
trated load  at  the  middle. 

The  above  investigation  shows  that  the  elastic  resilience  of 
a  beam  is  proportional  to  the  product  of  the  modulus  of  resilience, 
the  volume  of  the  beam,  and  the  ratio  (r/c)2.  Since,  however, 
this  ratio  always  has  a  numerical  value  which  is  the  same  for 
similar  sections,  it  may  be  stated  as  a  general  law,  that  the  resili- 
ences of  beams  having  similar  cross-sections  are  proportional  to 
their  volumes. 

The  strength  of  a  rectangular  beam  increases  with  the  square 
of  its  depth  and  its  stiffness  with  the  cube  of  the  depth  (Art.  56). 
The  resilience,  however,  increases  with  the  section  area.  Hence 
it  is  immaterial  whether  the  short  or  the  long  side  of  the  section 
is  placed  vertical  when  the  function  of  the  beam  is  the  resistance 
of  external  work. 

When  the  unit-stress  increases  from  S\  up  to  5,  the  resilience 
which  can  be  obtained  when  the  load  is  lessened  so  that  S  decreases 
to  Si  may  be  found  by  replacing  S2  in  the  above  formulas  by 
S2— Si2,  as  the  discussion  in  Art.  119  shows. 

Prob.  121a.  Deduce  an  expression  for  the  resilience  of  a  rectan- 
gular beam  fixed  at  both  ends  and  uniformly  loaded;  also  when  fixed 
at  one  end  and  supported  at  the  other. 

Prob.  1216.  Compute  the  horse-power  required  to  deflect,  50  times 
per  second,  a  wrought-iron  cantilever  beam,  2X3X72  inches,  so  that 
at  each  deflection  the  unit-stress  S  may  range  from  5  ooo  to  10  ooo 
pounds  per  square  inch. 


310  RESILIENCE  AND  WORK  CHAP,  xill 


ART.  122.     RESILIENCE  IN  SHEAR  AND  TORSION 

The  elastic  resilience  of  a  body  under  the  action  of  shear 
is  governed  by  similar  laws  to  that  of  tension  and  flexure,  namely, 

it  is  proportional  to  the  square  of 
the  maximum  unit-stress  and  to  the 
volume  of  the  body.  Thus  in  Fig. 
122  let  a  vertical  shear  act  upon  a 
parallelepiped  of  length  /  and  sec- 
tion area  a,  deforming  it  into  a 
rhombopiped.  The  figure  repre- 
sents a  short  beam  with  a  load  P 
at  the  end,  so  that  the  shear  in  every  vertical  section  is  equal  to 
P,  and  the  shearing  unit-stress  is  S8=P/a.  The  external  work 
done  by  the  load  P,  supposing  it  to  be  gradually  applied,  is  \Pe, 
where  e  is  the  distance  through  which  P  deflects.  This  work  is 
stored  in  the  body  in  the  form  of  stress  energy,  and  is  equal  to 
its  elastic  resilience.  Now  P=aS8  and  from  (15)  the  unit- 
detrusion  is  e=(Sa/F)l,  where  F  is  the  shearing  modulus  of 

elasticity.     Therefore 

K=$Pe  =  $(S82/F)al  (122) 

is  the  resilience  or  the  work  obtainable  from  the  stored  energy 
when  the  load  P  is  removed  provided  the  unit-stress  S8  is  not 
greater  than  the  shearing  elastic  limit  of  the  material. 

The  resilience  of  a  shaft  under  torsion  can  be  determined 
in  a  manner  similar  to  that  of  beams.  When  a  round  shaft  is 
twisted  by  a  force  P  acting  with  a  lever  arm  p,  as  in  Fig.  90, 
each  element  da  of  the  section  is  subject  to  a  shearing  unit-stress  S8 
and  to  a  total  stress  S8 .  da.  The  internal  energy  or  resilience 
for  this  element  is  then  \S8daXe,  where  e  is  the  deformation 
caused  by  the  shear.  Since  e  =  (Sa/F)lis  the  deformation  in 
the  distance  /,  the  internal  energy  stored  in  the  element  of  area 
da  and  length  dx  is, 

dK'  =  ±S.da  .  (Ss/F)l=%(S82/F)l .  da 

Now  let  S  be  the  shearing  unit-stress  at  the  circumference  most 
remote  from  the  axis,  and  let  c  be  its  distance  from  that  axis; 


ART.  122  RESILIENCE    IN    SHEAR   AND    TORSION  311 

also  let  z  be  the  distance  of  da  from  the  axis.  Then,  since  the 
stresses  vary  as  their  distance  from  the  axis,  S8  equals  S  .  z/c. 
Inserting  this  in  the  above  expression  and  integrating  it  over 
the  cross-section,  2 'da  .  z2  is  replaced  by  /,  the  polar  moment 
of  inertia,  or  by  ar2,  where  r  is  the  polar  radius  of  gyration.  Thus 
are  found, 

K  =  \(S2/F}U/c2  or  K=^(S2/F)(r/c)2al 

for  the  internal  energy  or  resilience  of  the  shaft.  Now  al  is 
the  volume  of  the  shaft,  and  it  is  thus  seen  that  the  resiliences 
of  circular  shafts  are  proportional  to  their  volumes.  Hence  the 
resilience  of  a  shaft  under  torsion  is  governed  by  laws  similar 
to  those  of  a  beam  under  flexure  (Art.  121). 

The  formula  here  established  is  only  valid  when  the  greatest 
unit-stress  S  does  not  surpass  the  elastic  limit  for  shearing. 
When  S  corresponds  to  the  elastic  limit,  the  quantity  $S2/F 
may  be  called  the  modulus  of  resilience  for  torsion  or  shearing, 
in  analogy  to  the  modulus  of  resilience  for  tension  or  compres- 
sion (Art.  120). 

As  an  example,  let  it  be  required  to  find  the  work  necessary 
to  stress  a  steel  shaft  12  inches  in  diameter  and  30  feet  long  up 
to  its  shearing  elastic  limit  of  30  ooo  pounds  per  square  inch. 
Here  5=30000  and  F  =  11200  ooo  pounds  per  square  inch 
(Art.  93);  also,  c=6  inches,  a  =  113.1  square  inches,  J  =  -fand4  = 
^2  =  2036  inches4,  7  =  360  inches.  Inserting  all  values,  K  is 
found  to  be  818000  inch-pounds  or  68200  foot-pounds;  to 
produce  this  stress  in  the  shaft  in  one  minute,  more  than  2  horse- 
powers are  required. 

The  resilience  of  a  shaft  under  torsion  is  measured  by  the  work 
required  to  produce  a  given  unit-stress  5,  and  from  the  above 
discussion  this  is  seen  to  vary  with  (r/cf.  To  compare  the 
resilience  of  a  hollow  shaft  of  outside  diameter  d  and  inside 
diameter  d\  with  that  of  a  solid  shaft  having  the  same  section 
area,  it  is  hence  only  necessary  to  compare  their  values  of  r2/c*t 
Let  d2  be  the  diameter  of  the  solid  shaft ;  then  for  the  two  cases, 

r2/c2  -  $d22/W  =  i 


312  RESILIENCE  AND  WORK  CHAP.  XIII 

and  accordingly,  representing  the  ratio  d\/d  by  /c,  there  is  found, 

hollow/solid  =  (/c2  +  1) 

which  is  the  ratio  of  the  resilience  of  a  hollow  shaft  to  a  solid 
one  of  the  same  section  area.  If  the  outer  diameter  of  the  hollow 
shaft  is  double  the  inner  diameter,  the  resilience  of  the  hollow 
shaft  is  25  percent  greater  than  that  of  the  solid  one. 

Prob.  1220.  Compare  the  resiliences  of  a  solid  shaft  16  inches  in 
diameter  with  that  of  a  hollow  shaft  18  inches  in  outside  and  8  inches 
in  inside  diameter. 

Prob.  122&.  A  simple  beam  of  section  area  a  and  span  /  has  a  load 
P  at  the  middle.  Show  that  the  resilience  due  to  the  vertical  shears 
isP2l/8aF. 

ART.  123.     DEFLECTION  UNDER  ONE  LOAD 

By  the  help  of  the  preceding  principles  regarding  work,  a 
method  will  now  be  established  by  which  the  internal  energy 
stored  in  a  beam  may  be  expressed  in  terms  of  the  bending  mo- 
ments. Using  the  same  notation  as  in  Arts.  40,  41,  the  horizontal 
unit-stress  at  any  distance  z  above  or  below  the  neutral  axis  is 
S  .  z/c.  In  the  horizontal  distance  dx  the  change  of  length  due 
to  this  unit-stress  is  by  (10)  known  to  be  (Sz/cE)dx.  The  ele- 
mentary work  of  a  fiber  of  area  da  under  the  gradually  applied 
unit-stress  hence  is  $(Sz/c)da(Sz/cE)dx.  Accordingly  the  work 
dK  stored  in  all  the  fibers  of  the  cross-section  in  the  distance  dx 
is  found  by  summing  the  works  of  all  the  fibers.  Noting  that 
Ida  .  z2  is  the  moment  of  inertia  /,  and  that  the  value  of  S/c 
is  M/I  from  the  flexure  formula  (41),  there  is  found, 


K- 


- 


2EI  .  J       2EI 

the  second  of  which  gives  the  total  stored  energy.  By  expressing 
the  bending  moment  M  as  a  function  of  x  and  integrating  the 
expression  over  the  entire  length  of  the  beam,  K  can  be  found 
for  any  particular  case. 

For  example,  consider  a  cantilever  beam  loaded  at  the  end 
with   P.     Here   M=—Px,    and   inserting   this   and   integrating 


ART.  123  DEFLECTION   UNDER   ONE   LOAD  313 

between  the  limits  o  and  /  gives  K=P213/6EI  for  the  total  stored 
energy  due  to  a  load  gradually  applied  at  the  end.  Again  con- 
sider a  cantilever  beam  loaded  uniformly  with  w  per  linear  unit; 
here  M=  —  %wx2,  whence  by  integrating  results  K  =  w2l5/4oEI; 
or  if  W  is  the  total  uniform  load,  K  =  W2P/4oEI,  which  is  ^  of 
that  caused  by  the  same  load  applied  at  the  end. 

The  formula  (123)  furnishes  a  new  and  convenient  method 
of  determining  the  elastic  deflection  of  a  beam  under  a  single  load 
P.  Let  /  be  the  deflection  under  the  load.  Then  \Pf  is  the 
external  work  done  by  P  as  it  gradually  increases  from  o  up  to 
the  value  P,  and  this  must  equal  the  stored  energy  K}  whence, 


or  Pf-  (123) 


gives  the  deflection  under  the  load.  For  example,  take  a  simple 
beam  loaded  at  the  middle  with  P;  the  value  of  M  is  iP#, 
and  the  total  integral  is  twice  the  integral  taken  between  the 
limits  o  and  J/.  Accordingly, 

/$/  p2^2^v  p/3 

•L         -V      ^*V  ,  /•  -t* 

whence  /  =• 

4£/ 


48E/ 

which  is  the  same  result  as  that  deduced  in  Art.  55  by  the  use 
of  the  equation  of  the  elastic  curve. 

The  deflection  of  the  metal  part  of  the  flitched  beam  of  Fig. 
1126,  when  the  length  of  the  metal  is  less  than  that  of  the  timber, 
can  readily  be  found  by  the  use  of  (123).  Let  /  be  the  span  of 
the  simple  beam  and  d  the  distance  from  each  end  of  it  to  the  end 
the  metal  plate.  Let  PI  be  the  single  load  at  the  middle  which 
is  borne  by  the  metal,  then  M  =  \P<zx,  but  in  the  integration 
the  limits  must  be  between  \l  and  d.  Let  /  be  the  moment  of 
inertia  of  the  metal  section  and  E  the  modulus  of  elasticity  of 
the  metal.  Then, 

P22x2dx  ,    P 

-&T        whence       /= 

is  the  deflection  of  the  metal  at  the  middle  of  the  beam,  and  the 
statement  made  in  the  last  paragraph  of  Art.  112  is  therefore 
justified. 

The  above  method  is  not  applicable  to  the  determination  of 


314  RESILIENCE  AND  WORK  CHAP.  XIII 

the  deflection  at  any  point  of  a  beam,  except  that  under  the  single 
load  P,  nor  can  it  be  used  for  several  single  loads  or  for  a  uniform 
load.  The  method  to  be  pursued  in  such  cases  is  developed  in 
the  next  article. 

Prob.  123.  Deduce,  by  the  above  method  and  also  from  Case  III 
of  Art.  55,  the  deflection  under  a  single  load  P  which  is  placed  on  a 
simple  beam  at  a  distance  J/  from  the  left  support. 

ART.  124.     DEFLECTION  AT  ANY  POINT 

Consider  a  beam  of  any  kind,  loaded  in  any  manner,  and 
let  it  be  required  to  find  the  deflection  at  a  given  point.  Let  a 
weight  P  be  imagined  to  be  put  at  this  point,  this  being  either 
a  part  of  the  weight  of  the  beam  concentrated  there  or  a  single 
V>ad  supposed  to  be  there.  The  deflection  at  this  point  due 
to  all  the  loads  on  the  beam  being  /,  the  work  of  the  weight  P 
is  JP/  and  this  must  equal  the  internal  energy  throughout  the 
beam  due  to  that  weight  P. 

Let  x  be  the  distance  of  any  section  of  the  beam  from  an 
assumed  origin,  S  the  unit-stress  on  the  extreme  fiber  of  this  sec- 
tion at  the  distance  c  from  the  neutral  axis,  and  (S  .  z/c]  da  the 
stress  on  a  fiber  of  area  da  at  a  distance  z  above  or  below  the 
neutral  axis,  both  of  these  being  due  to  all  the  loads  on  the  beam. 
Let  s  and  s  .  z/c  be  the  corresponding  unit-stresses  due  to  the 
weight  P  ;  the  change  of  length  in  the  fiber  of  area  da  in  the  hori- 
zontal distance  dx,  due  to  the  weight  P,  is  (sz/cE)dXj  and  the 
work  in  this  fiber  due  to  P  is  $(Sz/c)da(sz/cE)dx.  The  sum- 
mation of  all  values  of  this  product,  first  throughout  the  section 
and  secondly  throughout  the  length  of  the  beam,  gives  the  in- 
ternal energy  due  to  P.  Now  Ida  .  z2  is  the  moment  of  inertia 
/;  the  value  of  S/c  is  M/I,  where  M  is  the  bending  moment  due 
to  all  the  loads  on  the  beam,  and  the  value  of  s/c  is  M//,  where 
M  is  the  bending  moment  due  to  the  weight  P.  Accordingly, 
equating  the  external  work  to  the  internal  energy,  and  extending 
the  integration  over  the  length  of  the  beam,  there  results, 


or 


ART.  124  DEFLECTION   AT   ANY    POINT  315 

both  of  which  give  the  deflection  /  at  any  point  of  the  beam; 
in  the  first  M  is  the  bending  moment  due  to  any  weight  p  at  that 
point;  in  the  second  M'  is  the  bending  moment  due  to  the  weight 
unity  at  that  point.  By  inserting  for  M  and  M'  their  values  in 
terms  of  x,  and  integrating,  the  deflection  /  is  readily  found. 


Fig.  124a  Fig.  1246  Fig.  124c 

For  example,  let  it  be  required  to  find  the  deflection  at  the 
end  of  a  cantilever  beam  due  to  a  load  P  placed  at  a  distance 
K!  from  the  left  end,  as  in  Fig.  124a.  On  the  left  of  the  load 
M  =  o,  and  on  the  right  of  the  load  M=—P(x  —  id),  where  x 
is  measured  from  the  free  end.  Placing  a  weight  unity  at  the 
end,  the  bending  moment  due  to  it  is  M'  =  —  x  throughout  the 
entire  length  of  the  cantilever.  Thus  MM'=o  on  the  left  of 
the  given  load  and  MM'  =P(x2-Klx)  on  the  right.  The  formula 
then  furnishes, 

EIf=  f  P(x2-Klx)dx=$PP(2-3K+K3) 

JKl 

which  gives  the  deflection  of  the  end  due  to  P.     When  /c  =  o,  the 
load  is  at  the  end  and  /=  PP/EI,  as  otherwise  found  in  Art.  54. 


As  a  second  example,  let  it  be  required  to  find  the  deflection 
of  the  right  end  of  the  overhanging  beam  in  Fig.  1246,  the  load 
being  uniform  throughout.  Let  /  be  the  span  and  d  the  length 
of  the  overhanging  arm.  The  left  reaction  due  to  the  uniform 
load  is  \wl(i  -  K2),  whence  M  =  %wl(i  -  K2)x-  \wx2  for  the  span  /; 
for  the  overhanging  arm  let  x  be  any  distance  from  the  right  end, 
then  M  =  \wx2.  For  a  weight  unity  at  the  end  of  the  overhang- 
ing arm,  the  left  reaction  is  —  K,  whence  M'  =  —  nx  for  the  span  /; 
for  the  overhanging  arm  M'=x.  The  deflection  formula  then 
gives 

EIf=  - 


from  which/  is  known  for  any  value  of  K.     For  instance,  let  «=i, 
then/=  -  i  $wl4/2048EI,  the  minus  sign  showing  that  the  deflec- 


316  RESILIENCE  AND  WORK  CHAP,  xm 

tion  is  upward,  because  at  the  beginning  of  this  article  positive 
values  of  f  were  measured  downward.  For  a  longer  overhang- 
ing arm,  however,  /  may  be  positive  ;  this  is  the  case  when  K  =  J. 

When  a  cantilever  beam  deflects,  the  free  end  suffers  a  hori- 
zontal displacement  which  may  be  derived  from  (124),  taking 
the  force  unity  as  horizontal,  and  using  the  equation  of  the  elastic 
curve.  For  Fig.  124c,  where  the  load  P  is  at  the  end,  the  elastic 
curve  has  the  equation  ;y=/i(2/3  —  3/2#-f-#3)/2/3,  where  f\  is  the 
vertical  deflection  under  the  load  (Art.  54).  Now  let  the  hori- 
zontal force  unity  be  applied  to  the  free  end  where  the  horizontal 
displacement  is  /;  its  moment  is  M'  =  ~(f\—y)  and  the  moment 
of  P  is  M=  -Poc.  Then  the  formula  (124)  becomes, 

i  -?)***=  (Pf  i/  * 


from  which  the  horizontal  displacement  of  the  end  of  the  beam 
may  be  found  to  be  /=6/i2/5^  which  is  very  small  compared 
with  the  vertical  deflection  of  the  end. 

The  above  examples  show  the  great  value  of  formula  (124) 
in  discussing  all  questions  regarding  the  elastic  deflections  of 
beams.  It  may  be  used  to  find  the  equation  of  the  elastic  curve 
also,  taking  /  as  the  ordinate  corresponding  to  the  abscissa  x. 
It  applies,  however,  only  to  cases  where  the  greatest  flexural 
unit-stresses  do  not  exceed  the  elastic  limit  of  the  material. 
When  /  is  not  constant,  it  is  to  be  expressed  as  a  function  x 
before  the  integration  can  be  made. 

Prob.  124a.  Find  by  the  above  method  the  equation  of  the  elastic 
curve  for  a  simple  beam  uniformly  loaded,  and  compare  it  with  that 
deduced  in  Art.  55. 

Prob.  1246.  A  simple  beam  has  two  equal  concentrated  loads 
placed  at  equal  distances  from  the  supports.  Deduce  the  deflection 
under  one  of  the  loads,  and  also  the  deflection  at  the  middle  of  the 
beam. 

.  ART.  125.     DEFLECTION  DUE  TO  VERTICAL  SHEAR 

The  treatment  of  the  deflection  of  beams  in  the  previous 
pages  has  been  solely  from  the  standpoint  of  the  horizontal 


ART.  125  DEFLECTION   DUE    TO   VERTICAL   SHEAR  317 

stresses  caused  by  the  external  bending  moment.  Art.  122  shows, 
however,  that  internal  energy  is  stored  in  a  beam  by  a  vertical 
shear,  and  it  hence  appears  that  the  former  investigations  are 
more  or  less  incomplete  in  neglecting  the  deformation  which  is 
caused  by  the  external  vertical  shears. 

Referring  to  Fig.  125  let  V  be  the  vertical  shear  which  has 
produced  the  vertical  detrusion  df  in  the  short  length  dx.  Let  a 
be  the  section  area  of  the  beam,  F  the 
shearing  modulus  of  elasticity.  The 
shearing  unit-stress  in  the  section  is  V/a 
and  this  will  be  considered  as  uniformly 
distributed.  The  internal  work  in  the 
length  dx  is  JF  .  df  when  V  increases 
gradually  from  o  up  to  V.  Now  from 
the  definition  of  F  in  Art.  15  it  is  known  that  df=(V/aF)dx, 
since  df/dx  is  the  unit-detrusion  that  is  due  to  the  unit-shear  V/a. 
Accordingly  the  whole  internal  energy  due  to  shear  in  the  length 
dx  is  dK  =  J(  F2/  aF)  dx.  When  /  is  the  deflection  under  a  single 
load  P,  the  external  work  is  \Pf.  Accordingly, 


<- 505- •»• 


are  expressions  for  the  total  internal  energy  due  to  shearing, 
and  the  deflection  /  under  a  single  load  P. 

For  instance,  a  simple  beam  with  a  load  P  at  the  middle 
has  the  shear  F  =  JP.  Then  by  taking  double  the  integral 
between  the  limits  o  and  \l  there  is  found  K=  P2l/SFa  for  the 
internal  energy  or  resilience  due  to  the  shears  throughout  the 
beam.  The  deflection  under  the  load  due  to  the  vertical 
shear  then  is  f=Pl/^Fa.  In  Art.  55  the  deflection  due  10  the 
bending  moments  was  found  to  be  P13/4SEI.  Accordingly  the 
ratio  of  the  shearing  deflection  to  that  of  die  flexural  deflection 
is  i2EI/Fal2,  which  reduces  to  i2(E/F)(r/l)2,  where  r  is  the 
radius  of  gyration  of  the  section.  For  a  cast-iron  beam  the 
ratio  E/F  is  about  2.5;  if  the  section  is  square  the  ratio  of  the 
shearing  to  the  flexural  deflection  is  2.5(d//)2,  where  d  is  the  side 


318  RESILIENCE  AND  WORK  CHAP,  xill 

of  the  square;  when  the  length  of  the  square  beam  is  30  times 
its  side,  this  ratio  is  ^fo,  so  that  the  deflection  due  to  shearing  is 
small  compared  with  that  due  to  the  moments;  for  a  short  beam, 
such  as  l  =  2d,  this  ratio  is  0.625,  so  tnat  tne  deflection  due  to 
shearing  is  62 J  percent  of  that  due  to  the  moments. 

In  1870  Norton  called  attention  to  the  inaccuracy  of  the 
ordinary  formula  for  deflection  in  the  case  of  short  beams.  Experi- 
ments on  white-pine  beams  of  different  lengths  and  sizes  were 
made  with  loads  at  the  middle  of  the  spans,  and  it  was  shown 
that  the  deflections  were  directly  proportional  to  the  loads  and 
inversely  proportional  to  the  breadth  of  the  beam,  as  the  com- 
mon formula  requires.  The  deflections  were,  however,  not 
directly  proportional  to  the  cubes  of  the  spans  nor  inversely  pro- 
portional to  the  cubes  of  the  depths  of  the  beams,  as  the  formula 
requires.  An  examination  into  the  reason  of  these  discrepancies 
showed  that  it  was  due  to  influence  of  the  vertical  shears,  and 
Norton  showed  that  the  quantity  C .  Pl/bd  should  be  added 
to  the  common  flexure  formula  in  order  to  satisfy  the  results 
of  his  experiments;  he  further  showed  that  the  quantity  C  had 
the  value  o.ooo  0x3094  for  the  white-pine  beams.  From  the 
theoretic  value  deduced  in  the  last  paragraph  it  is  seen  that 
4F  is  the  reciprocal  of  this  number,  whence  F  =  266000  pounds 
per  square  inch,  which  should  be  the  modulus  of  elasticity  for 
the  shearing  of  white  pine  across  the  grain.  This  is  probably 
not  far  from  the  actual  value  of  that  coefficient,  since  Thurston, 
by  experiments  on  torsion,  found  jF  =  22oooo  pounds  per  square 
inch  for  white  pine.  The  experiments  of  Norton,  therefore, 
confirm  the  above  theory  of  the  deflection  of  beams  due  to 
shearing. 

Formula  (125)  does  not  apply  to  the  deflection  at  any  point, 
or  to  any  kind  of  loading  except  a  single  load  P.  For  any  other 
case  let  any  weight  P  be  at  the  point  where  it  is  desired  to  find  the 
deflection,  and  let  v  be  the  shear  due  to  this  weight.  The  external 
work  due  to  P  is  JP/,  where  fu  the  required  deflection.  The  in- 
ternal work  in  any  elementary  distance  dx  is  one-half  of  the  prod- 
uct of  the  actual  shear  Fand  the  detrusion  (v/aF)dx.  Therefore 


ART.  125  DEFLECTION    DUE   TO    VERTICAL    SHEAR  319 


/ 


or 


are  expressions  for  finding  the  deflection  due  to  shearing;  in  the 
first  v  is  the  shear  due  to  any  weight  P  and  in  the  second  V  is 
the  shear  due  to  the  weight  unity  at  the  point  where  the  deflec- 
tion is  desired. 

For  example,  take  the  case  of  Fig.  1240  and  let  it  be  required 
to  find  the  deflection  at  the  end  due  to  shearing  when  a  single 
load  P  is  at  the  distance  id  from  the  left  end.  Let  x  be  measured 
horizontally  from  the  free  end  where  the  weight  unity  is  placed. 
Then  V  =  o  on  the  left  of  the  load  and  V  =  —  P  on  the  right  of 
the  load,  while  V  =  —  i  throughout  the  beam.  Then  from  (125)', 

Faf=  flP3x=Pl(i-k)  or  f=(i-k)Pl/Fa 

JK! 

gives  the  deflection  at  the  end  due  to  the  vertical  shears.  When 
K=O,  the  load  P  is  at  the  end  and  f  =  Pl/Fa;  the  ratio  of  this 
to  the  deflection  due  to  the  moments  is  s(E/F)(r/l)2.  This 
shows  that  the  deflection  due  to  the  shears  is  scarcely  appreciable 
in  long  beams;  for  short  beams,  however,  it  may  be  larger  than 
that  due  to  the  moments. 

By  measuring  the  elastic  deflections  of  two  beams  of  different 
lengths  but  of  the  same  material,  it  is  possible  to  compute  the 
values  of  E  and  F  for  that  material.  The  deflections  thus  meas- 
ured are  due  both  to  moments  and  shears  and  hence  an  ex- 
pression for  each  measurement  is  to  be  written  in  terms  of  E 
and  F.  Let  l\  and  /2  be  the  spans,  #1  and  a2  the  section  areas, 
/i  and  /2  the  moments  of  inertia,  and  PI  and  P2  the  loads  at  the 
middle  of  the  simple  beams.  Let  the  measured  deflections  be 
/i  and  /2.  Then  may  be  written, 


l./Vi!.  P/23  i   ,P212  i  =f 

Jl  J 


48/1  £401  F  48/2  E      4^2  F 

which  contain  the  two  unknown  quantities  i/E  and  i/F,  and 
hence  the  solution  of  the  two  equations  will  furnish  the  values 
of  the  common  modulus  E  and  the  shearing  modulus  F.  By 
making  many  experiments  instead  of  two,  writing  an  equation 
for  each,  and  solving  the  equations  by  the  method  of  Least  Squares, 


320  RESILIENCE  AND  WORK  CHAP,  xin 

it  is  possible  to  obtain  reliable  values  of  these  moduluses  of  elas- 
ticity. In  such  experiments  it  is  of  course  important  that  the 
loads  should  not  be  heavy  enough  to  stress  the  material  beyond 
its  elastic  limit. 

In  conclusion  it  may  be  noted  that  there  is  an  elastic  curve 
due  to  vertical  shears,  and  its  equation  may  be  used  in  similar 
manner  to  that  of  the  bending  moments.  From  Fig.  125,  the 
general  equation  of  this  elastic  curve  is, 

£-4       or       af-v 

ox     ab  ox 

For  example,  let  it  be  required  to  find  the  elastic  curve  and  the 
deflection  due  to  vertical  shears  for  a  simple  beam  of  span  / 
loaded  uniformly  with  w  per  linear  unit.  Here  V  =  \wl— wx, 
and  inserting  this  in  the  general  equation  and  integrating  there 
is  found  aFy  =  ^w(lx  —  x2)  for  the  equation  of  the  elastic  curve. 
The  maximum  deflection  occurs  at  the  middle  where  x=%l,  and 
its  value  /  =wl2/SaF}  and  this  agrees  with  the  result  which  will 
be  found  by  the  use  of  (125)'. 

Prob.  125a.  Find  the  deflection  of  the  end  of  the  overhang  in 
Fig.  1246  due  to  the  vertical  shears  of  the  uniform  load. 

Prob.  I25b.  A  timber  beam  9  inches  long  and  2  inches  square  is 
placed  on  supports  7  inches  apart  and  subject  to  deflection  by  a  load 
of  300  pounds  at  the  middle.  Compute  the  amount  of  the  deflection 
due  to  the  vertical  shears  and  that  due  to  the  bending  moments. 

ART.  126.     PRINCIPLE  OF  LEAST  WORK 

There  is  a  certain  law  of  nature  called  the  principle  of  least 
work,  which  is  of  great  value  in  discussing  problems  in  mechanics 
that  cannot  be  solved  by  static  conditions  alone.  For  example, 
a  table  with  four  legs  is  a  system  where  the  reactions  of  the  legs, 
due  to  an  unsymmetric  load  on  the  table,  cannot  be  found  by 
statics,  since  the  three  conditions  of  static  equilibrium  cannot 
determine  four  unknown  quantities.  The  principle  of  least  work 
furnishes,  however,  a  fourth  condition  when  the  stresses  in  the 
legs  do  not  exceed  the  elastic  limit  of  the  material. 


ART.  126  PRINCIPLE    OF    LEAST    WORK  321 

When  a  structure  is  so  formed  that  the  stresses  in  it  cannot  be 
determined  by  statics,  it  is  said  to  be  a  "  redundant  system",  and 
the  principle  of  least  work  applicable  to  it  is  as  follows: 

The    stresses    in  a  redundant  system  have   such    values 
that  the  internal  energy   of   all  the  stresses  is  a  minimum. 

This  is  sometimes  regarded  as  an  axiom,  it  being  considered  that 
the  resisting  forces  will  store  up  no  more  energy  than  the  minimum 
which  is  necessary  to  maintain  equilibrium  with  the  external 
forces.  The  following  proof,  however,  will  probably  be  more 
satisfactory  than  the  assumption  of  the  axiom.  Let  PI,  P2,  .  .  .  Pr 
be  r  forces  in  equilibrium,  and  let  a  small  displacement  be  made, 
without  performing  work  on  the  system,  so  that  the  points  of 
application  of  these  forces  move  through  the  small  distances 
o<?i,  de2,  .  .  .  der.  Then  t'he  total  work  done  must  be  equated  to 
zero,  or,  Pl(^1  +  p2^2-fp3&>3+  .  .  .  +prder=o 

Now  the  condition  that  this  equals  zero  is  the  same  as  the  condi- 
tion that  its  integral  shall  be  a  minimum,  or 

PI&I  +  P2e2+ -^3^3+  •  •  •  +Prer  =  a.  minimum 

in  which  e\t  e2,  ...  er  are  the  total  distances  through  which  the 
forces  have  moved  in  acquiring  the  state  of  stable  equilibrium. 
Now  each  of  the  quantities  Pe  is  double  the  work  done  by  the 
applied  force  P  as  it  increases  from  o  to  P,  and  hence  also  equal 
to  double  the  energy  stored  by  the  stresses  that  balance  it.  There- 
fore, the  internal  work  or  stored  energy  of  all  the  resisting  stresses 
is  a  minimum. 

When  forces  act  upon  elastic  bodies,  in  which  the  defor- 
mations are  proportional  to  the  stresses,  the  above  principle  of 
least  work  may  be  applied  to  determine  unknown  reactions 
when  the  conditions  of  statics  are  not  sufficient  in  number  to 
do  so.  Many  problems  of  fixed  and  continuous  beams  may  be 
discussed  by  the  aid  of  this  principle.  For  example,  take  a  con- 
tinuous beam  of  two  equal  spans  loaded  uniformly  in  the  second 
span,  as  in  Fig.  126a,  and  let  it  be  required  to  find  the  three  reac- 
tions Rij  R2,  RS.  The  first  condition  of  statics  is  that  the  sum 
of  the  reactions  equals  the  total  load,  or  ^1+^2  +  ^3=^;  the 


322  RESILIENCE  AND  WORK  CHAP.  Xlii 

second  condition  is  that  the  sum  of  the  moments  of  the  reactions 
equals  the  moment  of  the  load,  whence,  for  an  axis  at  the  right 
end,  2Ril  +  R2l  =  %Wl.  The  third  condition  is  that  the  internal 
energy  of  all  the  flexural  stresses  shall  be  a  minimum.  For  the 
first  span  the  bending  moment  at  the  distance  x  from  the  left 
support  is  M=R\x\  for  the  second  span  the  moment  at  the 
distance  x  from  the  right  support  is  M  =  R3x  —  fyw  x2.  Then, 
by  Art.  123,  the  total  internal  energy  of  the  horizontal  flexural 
stresses  is, 


Differentiating  this,  and  equating  the  .derivative  to  zero,  gives 
the  equation  SRi  .dRi  +  SR3.dR3-^W  .dR3  =  o  as  the  condition 
of  least  work.  Solving  the  three  equations  thus  established, 
there  result  RI  =  -  £>W,  R2  =  +  f  W  ,  A*  =  +-&W  for  the  reactions 
due  to  the  given  load;  these  are  the  same  results  as  derived  by 
the  use  of  the  theorem  of  three  moments  (Art.  71). 


A  A  A  A 

12  34 


Fig.  126a  Fig.  1266 

As  a  second  example,  take  the  partially  continuous  girder 
in  Fig.  126Z>  which  has  four  supports  and  a  joint  at  the  middle 
of  the  second  span  so  that  the  bending  moment  there  is  always 
zero.  Let  2/  be  the  length  of  the  middle  span  and  /  that  of  each 
end  span,  and  let  it  be  required  to  find  the  four  reactions  due 
to  a  uniform  load  in  the  last  span.  Three  conditions  are, 


the  first  being  the  static  condition  that  the  sum  of  the  vertical 
forces  is  zero;  the  second  the  static  condition  that  the  sum  of 
the  moments  of  these  forces  is  zero,  the  axis  being  taken  at  the 
right  end;  the  third  the  condition  that  there  is  no  moment  at 
the  joint.  From  these  three  conditions  the  values  of  three  reac- 


ART.  126  PRINCIPLE   OF    LEAST   WORK  323 

tions  may  be  found  in  terms  of  the  other  reaction,  thus, 


and  the  value  of  RI  may  be  found  by  the  help  of  the  principle 
of  least  work.  To  do  this  an  expression  for  the  bending  moment 
is  found  for  each  of  the  four  parts  of  the  beam  and  the  sum  of 
all  the  values  of  M2/2EI  will  give  the  total  internal  work  of  the 
flexural  stresses  (Art.  123).  Thus,  in  a  manner  similar  to  that 
of  the  last  paragraph,  a  fourth  condition  is  established  which 
expresses  that  the  work  of  the  internal  stresses  is  a  minimum. 
This  equation,  in  connection  with  the  three  previously  estab- 
lished, gives  Ri  =  -&W,  R2  =  +&W,  R3  =  +&W,  R*  =  +  ^W. 
These  are  the  reactions  for  the  given  uniform  load  due  to  the 
bending  moments,  and  they  will  be  slightly  modified  if  the  ver- 
tical shears  are  taken  into  account. 

Examples  of  the  application  of  the  principle  of  least  work  to 
bridges  and  arches  will  be  found  in  Parts  I  and  IV  of  Roofc 
and  Bridges.  This  principle  must  be  used  with  caution  by  tht 
beginner,  but  it  is  one  of  much  value  in  the  discussion  of  struc- 
tures which  are  statically  indeterminate. 

Prob.  126.  A  table  of  length  /  and  width  h  has  four  legs  at  the 
corners  which  are  of  equal  size  and  length.  For  a  load  P  placed  on 
the  table  at  a  distance  \l  from  the  side  h  and  a  distance  \k  from  the 
side  /,  compute  the  reactions  of  the  leg  nearest  the  load. 


324  IMPACT  AND  FATIGUE  CHAP,  xiv 


CHAPTER  XIV 

IMPACT  AND  FATIGUE 
ART.  127.     SUDDEN  LOADS  AND  STRESSES 

A  load  at  rest  on  a  bar  or  beam  is  called  a  '  static  load  ', 
The  same  term  is  applied  to  a  load  which  increases  from  o  up 
to  its  final  value  P  in  such  a  way  that  the  deformations  of  the 
bar  or  beam  at  different  instants  are  proportional  to  the  loads 
aciing  at  those  instants  until  the  elastic  limit  of  the  material  is 
exceeded.  Loads  applied  in  any  other  manner  are  sometimes 
called  '  dynamic  ',  and  the  term  "  impact  "  implies  either  sud- 
denness of  action  or  that  the  load  is  in  motion  before  it  is  applied 
tc  the  bar  or  beam.  Static  loads  have  been  mostly  considered 
in  the  preceding  chapters,  but  it  has  always  been  recognized 
that  the  stresses  and  deformations  due  to  sudden  and  variable 
loads  are  greater  than  for  static  ones  (Art.  7).  The  terms 
dynamic  stress  '  and  '  dynamic  deformation  '  are  sometimes 
ttsed  to  distinguish  the  effects  of  impact  from  those  due  to  static 
loads. 

A  static  tensile  load  is  usually  applied  to  a  bar  by  increments, 
so  that  it  increases  from  o  up  to  P  in  such  a  way  that  the  elonga- 
tion is  proportional  to  the  load  until  the  elastic  limit  of  the  mate- 
rial is  reached.  The  work  done  upon  the  bar  is  then  equal 
to  the  mean  load  \P  multiplied  by  the  elongation  e,  or  K  =  ^Pe. 
Simultaneously  the  stress  in  the  bar  increases  from  o  up  to  P 
and  the  internal  energy  stored  in  the  bar  is  ^Pe.  The  triangle 
in  Fig.  127a  represents  both  the  external  work  and  the  internal 
energy. 

When  a  load  is  applied  to  a  bar  in  such  a  manner  that  its 
intensity  is  the  same  from  the  beginning  to  the  end  of  the  elonga- 
tion, it  is  called  a  "sudden  load  ".  For  instance,  let  a  load  be 
hung  by  a  cord  and  just  touch  a  scale-pan  at  the  foot  of  a  vertical 
bar;  then  if  the  cord  is  quickly  cut,  the  load  acts  upon  the  bar 
with  uniform  intensity  throughout  the  entire  elongation.  In 


ART.  127 


SUDDEN  LOADS  AND  STRESSES 


325 


this  case  the  maximum  elongation  is  greater  than  for  a  static 
load,  but  the  bar  at  once  springs  back,  carrying  the  load  with 
it,  and  a  series  of  oscillations  ensues,  until  finally  the  bar  comes 
to  rest  with  an  elongation  due  to  the  static  load.  Here  the  stress 
in  the  bar  increases  from  o  up  to  Q,  the  stress  Q  being  equal 
to  the  static  load  which  would  produce  the  maximum  elongation. 
Fig.  1276  represents  this  case,  where  the  rectangle  shows  the 
work  done  by  the  instantaneous  load  P,  and  the  triangle  shows 
the  internal  energy  stored  in  the  bar  at  the  instant  of  greatest 
elongation.  The  unit-stress  due  to  Q  must  be  less  than  the  elastic 
limit  in  order  that  the  following  discussions  may  be  valid. 


Fig.  127a 


—£—- 


Fig.  1276 


Fig.  I27c 


Let  q  be  the  maximum  elongation  due  to  the  sudden  load  P; 
the  work  performed  in  the  bar  is  Pq.  The  internal  energy 
stored  in  the  bar  at  the  instant  of  greatest  elongation  is  %Qq, 
since  the  stress  increases  from  o  up  to  Q.  Hence  %Qq=Pq,  or 
Q  =  2P.  Let  e  be  the  elongation  due  to  the  static  load  P;  then 
q/e  =  Q/P,  and  hence  also  q=2e.  Accordingly  the  following 
important  law  is  established  for  a  bar  under  elastic  changes  of 
length : 

A  sudden  load  produces  double  the  stress  and  double  the 
deformation  that  is  caused  by  a  static  load. 

Fig.  1300  shows  the  end  of  a  bar  acted  upon  by  a  sudden  load 
and  it  will  be  explained  in  Art.  132  that  the  maximum  velocity 
of  the  load  occurs  when  the  elongation  is  equal  to  e\  the  curve 
in  the  figure  shows  the  variations  in  velocity.  When  the  elonga- 
tion ze  is  reached,  the  stress  in  the  bar  is  2P  and  the  resultant 
force  tending  to  move  the  end  is  P-2P  or  -P;  hence  the  end 
moves  backward  and  oscillations  ensue,  until  finally  the  bar  comes 
to  rest  under  the  elongation  e  and  the  stress  P. 


326  IMPACT  AND  FATIGUE  CHAP.  XIV 

In  the  above  discussion  P  and  Q  are  the  total  stresses  on 
the  section  area  a  of  the  bar.  Let  S  and  T  be  the  corresponding 
unit-stresses,  so  that  P  =  Sa  and  Q  =  Ta.  Then  the  equation 
Q  =  2P  becomes  T  =  2S,  that  is,  the  unit-stress  due  to  a  sudden 
load  is  double  that  due  to  the  same  static  load. 

Similar  conclusions  result  when  a  single  load  P  is  suddenly 
applied  to  a  beam  producing  the  deflection  q  while  under  the 
same  static  load  the  deflection  is  /.  Let  Q  be  the  static  load 
which  will  produce  the  deflection  q.  Then  the  external  work  of 
this  static  load  is  %Qq,  while  that  of  the  sudden  load  is  Pq',  hence 
}Qq  =  Pq  or  Q  =  2P\  that  is,  the  sudden  load  P  produces  the 
same  effect  as  a  static  load  2P.  From  the  static  law  Q/P  =  q/f, 
it  follows  that  q  =  2f,  so  that  the  dynamic  deflection  is  double 
the  static  deflection.  Let  S  be  the  flexural  unit-stress  at  the 
dangerous  section  of  the  beam  when  the  deflection  is  /  and  let 
T  be  the  unit-stress  when  the  deflection  is  q;  then  q/f=T/S, 
since  elastic  deflections  are  proportional  to  the  unit-stresses 
(Art.  2).  Therefore  also,  T  =  2S,  that  is,  the  flexural  unit-stress 
due  to  a  sudden  load  is  double  that  due  to  the  same  static  load. 

Lastly,  consider  a  bar  upon  which  rests  a  load  PI  causing  the 
elongation  e\.  Let  a  sudden  load  P  be  now  brought  upon  it 
causing  the  additional  elongation  q  and  the  additional  stress  Q. 
Fig.  127 c  represents  this  case  and  it  shows  that  the  elongation  is 
ei  +  2e  and  that  the  final  stress  is  PI  +  2P;  thus  the  instantaneous 
load  produces  its  effect  independently  of  the  other.  As  soon  as 
the  elongation  e\  +  2.e  occurs,  the  bar  springs  back,  and  a  series 
of  oscillations  follows;  finally  the  bar  comes  to  rest  under  the 
elongation  ei+e  and  the  stress  P\+P.  Similar  conclusions  fol- 
low in  the  case  of  a  beam  under  an  initial  load. 

In  the  above  investigation  it  has  been  supposed  that  all  the 
work  Pq  performed  by  the  sudden  load  P  is  expended  in  storing 
energy  in  the  bar  or  beam.  This  is  not  strictly  the  case,  for 
internal  molecular  friction  causes  a  slight  loss  of  work  (Art.  147). 
The  law  deduced  is,  however,  very  close  for  a  light  beam,  but  Q 
is  really  a  little  less  than  2?  and  q  is  a  little  less  than  26  when  the 
beam  is  heavy  compared  with  the  load. 


ART.  128  AXIAL   IMPACT   ON  BARS  327 

Prob.  127.  A  vertical  steel  bar,  2  inches  in  diameter  and  13  feet 
long,  has  a  load  of  15  ooo  pounds  hung  at  its  end.  Compute  the 
elongation  due  to  this  static  load,  and  also  the  maximum  elongation 
which  occurs  when  an  additional  load  of  7  500  pounds  is  suddenly 
applied. 

ART.  128.    AXIAL  IMPACT  ON  BARS 

The  word  impact  is  here  used  to  mean  the  effect  of  a  load 
which  is  moving  when  it  strikes  the  end  of  a  bar;  such  a  load 
evidently  produces  a  greater  deformation  and  a  greater  stress 
than  one  applied  suddenly.  The  stress  in  the  bar  increases 
from  o  up  to  a  certain  limit  Q  and  the  deformation  increases 
from  o  up  to  q.  If  the  elastic  limit  of  the  material  is  not  ex- 
ceeded, the  stress  at  any  instant  is  proportional  to  the  deforma- 
tion so  that  the  stored  energy  of  the  internal  stresses  is  \Qq. 
Equating  this  to  the  external  work,  the  values  of  Q  and  q  may 
be  found. 

Let  P  be  a  weight  which  is  moving  horizontally  with  the  velocity 
V  at  the  instant  it  strikes  the  end  of  a  light  horizontal  bar.  Its 
kinetic  energy  is  P  .  V2/2g,  where  g  is  the  acceleration  of  gravity; 
or,  if  h  is  the  height  through  which  P  has  fallen  to  acquire  the 
velocity  F,  then  V2/2g  =  h,  and  the  kinetic  energy  of  the  moving 
weight  is  P  .  h.  Accordingly  %Qq  =  Ph,  if  no  work  is  expended 
in  overcoming  inertia  or  in  friction.  Now,  let  e  be  the  elongation 
of  the  bar  due  to  the  static  load  P ;  then  the  law  of  proportionality 
gives  q/e  =  Q/P.  From  these  two  equations  are  found, 

Q  =  P(2h/e)*  q  =  e(2h/e)*  =  (2he)*  (128) 

which  shows  that  Q  may  be  much  greater  than  P  and  q  much 
greater  than  e  when  the  weight  P  is  moving  rapidly.  For  example, 
let  F  =  io  feet  per  second,  then  h  =100/2X32. 2  =  1.55  feet  =  i8.6 
inches.  Let  the  weight  P  be  60  pounds  and  the  horizontal  bar 
be  of  steel  18  feet  long  and  3  square  inches  in  section  area,  then 
(10)  gives  e=Pl/aE=o.ooo  144  inches.  Accordingly,  6  =  3050x3 
pounds  and  5  =  0.073  inches,  which  are  about  510  times  as  great 
as  those  due  to  a  static  load  of  60  Ibs. 

When  a  vertical  bar  is  subject  to  the  impact  of  a  falling  weight 


328  IMPACT  AND  FATIGUE  CHAP  XIV 

P,  the  end  of  the  bar  is  elongated  or  shortened  the  amount  q 
so  that  the  work  performed  upon  it  is  (P(h+q).  The  internal 
stored  energy  is  %Qq  as  before.  Accordingly  the  two  equations 
iQq=p(-h  +  q)  anc[  q/e  =  Q/P  are  to  be  used  to  find  the  values 
of  Q  and  q,  which  are, 

<2  =  P+P(i  +  2/00*  q  =  e+e(i.  +  2h/e)*  (128)' 

When  h  =  o,  these  formulas  reduce  to  Q  =  2P  and  q  =  2e,  which 
are  the  results  previously  found  for  a  sudden  load.  Since  e  is  a 
small  quantity,  it  follows  that  a  load  dropping  from  a  moderate 
height  may  produce  large  stresses  and  deformations.  Experiments 
made  upon  elongations  of  spiral  springs  give  results  which  closely 
agree  with  those  computed  from  the  formula  for  q,  when  the 
elastic  limit  is  not  surpassed  by  the  stress  Q.  The  curve  in 
Fig.  1306  shows  the  variation  in  velocity  of  the  end  of  the  bar. 

The  effect  of  loads  applied  with  impact  is  therefore  to  cause 
stresses  and  deformations  greatly  exceeding  those  produced  by 
the  same  static  loads,  so  that  the  elastic  limit  may  perhaps  be 
often  exceeded.  Moreover  the  rapid  oscillations  which  ensue 
cause  a  change  in  molecular  structure  which  impairs  the  elasticity 
of  the  material  when  such  loads  are  often  applied.  It  is  some- 
times found  that  the  appearance  of  a  fracture  of  a  bar  which  has 
been  subject  to  shocks  is  of  a  crystalline  nature,  whereas  the 
same  material,  if  ruptured  under  a  gradually  increasing  stress, 
would  exhibit  a  tough  fibrous  structure.  Moving  loads  which 
produce  stresses  above  the  elastic  limit  cause  the  wrought  iron 
and  steel  to  become  stiff  and  brittle,  and  hence  it  is  that  the  work- 
ing unit-stresses  should  be  taken  very  low  (Art.  7). 

The  above  formulas  apply  also  to  unit-stresses.  Let  a  be 
the  section  area  of  the  bar,  S  the  unit-stress  due  to  the  static 
load  P  and  T  the  unit-stress  due  to  Q,  so  that  S=P/a  and  T  = 
Q/a.  Then  the  formulas  for  Q  in  (128)  and  (128)'  become, 


the  first  of  which  applies  to  a  horizontal  bar  and  -the  second  to  a 
vertical  bar.  For  instance  let  h=i8e,  then  T  =  6S  for  the  hori- 
zontal and  T  =  j.oSS  for  the  vertical  bar.  Here  T  is  the  unit- 


ART.  129  IMPACT   ON   BEAMS  329 

stress  which  prevails  in  the  bar  at  the  instant  of  greatest  defor- 
mation, but  after  a  series  of  oscillations  the  bar  comes  to  rest  under 
the  unit-stress  S.  These  oscillations  are  discussed  in  Art.  132. 

All  of  the  above  formulas  for  dynamic  stress  and  elongation 
give  results  which  are  somewhat  too  large,  because  a  portion  of 
the  energy  of  the  moving  weight  P  is  expended  in  overcoming 
the  inertia  of  the  bar.  They  apply  only  to  bars  which  are  so 
light  that  this  resistance  of  inertia  may  be  disregarded.  In 
Art.  130  it  will  be  shown  how  these  formulas  may  be  modified 
so  as  to  take  into  account  the  inertia  of  the  bar. 

Prob.  128.  In  an  experiment  upon  a  spring,  a  static  load  of  14.79 
ounces  produced  an  elongation  of  0.42  inches,  but  when  dropped 
from  a  height  of  7.72  inches  it  produced  a  stress  of  102.3  ounces  and 
an  elongation  of  2.90  inches.  Compare  theory  with  experiment. 


ART.  129.    IMPACT  ON  BEAMS 

When  a  falling  weight  strikes  a  beam,  it  causes  a  greater 
deflection  than  a  load  suddenly  applied.  Let  the  weight  P 
fall  from  a  height  h  above  a  light  beam  and  produce  the  dynamic 
deflection  </;  the  work  performed  is  then  P(h  +  q).  Let  T  be 
the  maximum  flexural  unit-stress  produced  by  the  impact  and 
S  be  that  due  to  a  static  load  P  which  causes  the  deflection  /. 
Then  the  deflections  are  proportional  to  the  unit-stresses,  if 
the  elastic  limit  is  not  exceeded,  or  q/f=T/S.  Also  let  Q  be  a 
static  load  which  will  produce  the  deflection  q  ;  then  the  deflec- 
tions are  also  proportional  to  the  loads,  or  q/f=Q/P]  accord- 
ingly Q/P  =  T/S.  The  external  work  of  the  load  (Ms  %Qq  and 
this  is  equal  to  the  internal  energy  stored  in  the  beam  when  the 
deflection  q  is  attained,  if  all  the  work  is  expended  in  stressing 
the  beam.  Hence  %Qq  =  P(h  +  q),  which  by  the  above  ratio 
reduces  to  %Tq=S(h+qT.  Combining  this  with  q/f=T/S,  there 
are  found, 

(129) 


as  the  formulas  for  the  dynamic  maximum  unit-stress  and  deflec- 
tion due  to  the  impact  of  a  single  load  P.     Here  S  is  found  from 


330  IMPACT  AND  FATIGUE  CHAP.  XIV 

the  flexure  formula  (41)  for  any  given  case,  or  S=Pcl/aI,  and 
/  is  found  from  the  deflection  formula  f=Pl3/ftEI,  where  /  is 
the  length  of  the  beam,  a  and  f)  numbers  depending  on  the 
arrangement  of  the  ends,  /  the  moment  of  inertia  of  the  cross- 
section,  c  the  distance  from  the  neutral  axis  to  the  remotest  fiber 
of  the  dangerous  section,  and  E  the  modulus  of  elasticity  of  the 
material  (Arts.  55  and  63). 

When  a  weight  P  is  moving  with  the  velocity  V,  it  can  per- 
form in  coming  to  rest  the  work  P  .  V2/2g,  where  g  is  the  accelera- 
tion of  gravity.  When  the  weight  moves  horizontally  and  strikes 
normally  against  the  side  of  a  beam  which  has  its  ends  arranged 
so  as  to  prevent  lateral  motion,  a  lateral  dynamic  deflection 
results.  Let  h  be  the  height  corresponding  to  V2/2g,  then  the 
~xternal  work  Ph  is  to  be  equated  to  %Qq  as  before.  Hence 
tne  equations  %Qq  =  Ph,  in  connection  with  the  laws  of  propor- 
tionality, give, 

T=S(2h/f)*  q=f(2h/f)*  =  (2hf)*  (129)' 

for  the  unit-stress  and  lateral  deflection  at  the  instant  when 
P  comes  to  rest.  This  case  rarely  occurs  except  in  machines 
for  testing  materials  by  impact. 

The  above  formulas  are  only  valid  when  the  unit-stress  7 
is  less  than  the  elastic  limit  of  the  material.  When  the  load  P 
is  light  compared  to  the  weight  of  the  beam,  they  give  results 
which  are  somewhat  too  large,  because  a  part  of  the  work  due 
to  P  is  expended  in  overcoming  the  inertia  of  the  beam  (Art.  131). 
It  will  be  noted  that  these  formulas  are  the  same  as  those  found 
in  Art.  128  for  bars,  except  that  the  static  deflection  /  appears 
instead  of  the  static  elongation  e. 

Prob.  129tf.  A  simple  beam  of  steel,  1X1X24  inches,  was  loaded 
with  a  weight  of  25  pounds  at  the  middle  and  the  deflection  found  to 
be  0.0028  inches.  It  was  then  struck  horizontally  by  a  hammer  weigh- 
ing 25  pounds  which  had  a  vertical  fall  of  2  inches.  Compute  the 
lateral  dynamic  deflection. 

Prob.  129&.  Compute  the  deflection  of  the  above  beam  when  a 
weight  of  25  pounds  falls  vertically  upon  the  middle  through  a  height 
*/f  2  inches.  The  observed  deflection  in  this  case  being  0.130  inches, 
what  explanation  may  be  given  of  the  smaller  computed  deflection? 


ART.  130  INERTIA  IN  AXIAL  IMPACT  331 


ART.  130.    INERTIA  IN  AXIAL  IMPACT 

When  a  moving  weight  strikes  axially  upon  the  free  end  of 
a  bar,  some  of  its  kinetic  energy  is  expended  in  overcoming  the 
inertia  of  the  particles  and  putting  them  into  motion,  this  energy 
being  converted  into  heat.  The  load  P  falling  through  the 
height  h  has  the  kinetic  energy  Ph  when  it  touches  the  end  of 
the  bar,  but  owing  to  the  loss  in  impact  only  a  part  of  Ph  is  effective 
in  elongating  and  stressing  the  bar.  Let  -n  be  a  number  less 
than  unity,  called  the  '  inertia  coefficient ',  then  t]Ph  is  the  energy 
which  produces  the  stress  Q  and  the  elongation  q  in  the  bar. 
All  the  formulas  of  Art.  128  may  hence  be  applied  to  heavy 
bars  when  the  number  y  is  known  by  replacing  h  by  yh.  The 
object  of  this  article  is  to  determine  the  value  of  y  in  terms  of  P 
und  the  weight  W  of  the  bar.  The  theory  of  the  impact  of 
inelastic  bodies  may  be  used  for  this  purpose  with  close  approxi- 
mation, since  the  moving  weight  and  the  end  of  the  bar  are  in 
close  contact  during  the  period  of  the  impact.  The  velocity 
with  which  stress  is  propagated  through  the  bar  will  be  sup 
posed  to  be  infinite.  The  greatest  unit-stress  Q/a,  where  a  is 
the  section  area,  must  not  exceed  the  elastic  limit  of  the  material. 

As  soon  as  the  load  P  strikes  the  end  of  the  bar,  its  velocity  V 
decreases  and  the  end  of  the  bar  begins  to  move.  When  com- 
plete contact  is  attained,  both  P  and  the  end  of  the  bar  are  moving 
with  a  velocity  v  which  is  less  than  F,  and  at  this  instant  any 
element  dW  of  the  bar  is  moving  with  a  velocity  u.  Accordingly 
the  kinetic  energy  stored  in  the  load  P  and  in  the  bar  of  length 
/  and  weight  W  at  this  instant  is  expressed  by, 


K  =  P.V2/2g  +         dW.U2/2g 
t/0 

Now  u=o  for  the  fixed  end  and  u=v  for  the  free  end  of  the 
bar,  and  for  an  infinite  velocity  of  stress  u  is  proportional  to 
the  distance  y  from  the  fixed  end,  so  that  u=v.y/l;  also  the 
element  dW  is  W  .  dy/l.  Introducing  these  values,  the  integral 
in  the  above  expression  is  found  to  be  %W .  v2/2g  or  one-third 
of  the  kinetic  energy  which  would  obtain  if  the  entire  bar  were  in 


332  IMPACT  AND  FATIGUE  CHAP  xiv 

motion  with  the  velocity  v.     Accordingly, 

K=(P+ %W)v2/2g  or  K=(i  +  $W/P.  P  .  v2/2g 

is  the  kinetic  energy  in  the  load  and  bar  when  the  load  and  the 
end  of  the  bar  are  moving  with  the  velocity  v. 

Now  it  is  known  from  Newton's  second  law  of  motion,  that 
the  common  velocity  of  two  bodies  at  the  instant  of  complete 
contact  in  the  impact  is  v  =  V  .P/(P+Pi),  when  the  body  of 
weight  P  moving  with  the  velocity  V  strikes  a  free  body  of  weight 
PI  which  is  at  rest.  For  the  case  in  hand,  however,  one  end  of 
the  bar  is  fixed,  so  that  W  cannot  replace  PI  in  this  expression. 
When  the  free  end  of  the  bar  is  moving  with  the  velocity  v,  the 
element  dW  at  the  distance  y  from  the  fixed  end  is  moving  with 
the  velocity  u=v.y/l  if  stress  is  transmitted  instantaneously. 
Accordingly,  instead  of  PV  =  (P  +  Pi)v  there  must  be  written, 


//  TT7"  /"*  7 

dW  .  u  =  Pv+-j£  I    ydy 


and  hence  v  =  V/(i+%W/P)  is  the  common  velocity  of  the  load 
and  the  end  of  the  bar.     Hence,  if  h  is  the  height  V2/2g,  the  above 
expression  for  K  becomes 
„       i+$W/P    . 

= 


in  which  y  is  the  inertia  coefficient.  When  the  bar  has  no  weight, 
then  T)  =  i  and  the  entire  kinetic  energy  is  effective  in  elongating 
and  stressing  the  bar.  When  the  load  and  bar  are  of  equal 
weight,  then  i?=Jf.  so  that  H-P^  is  effective,  while  j\Ph  is 
lost  in  heat. 

For  the  case  of  horizontal  impact  in  Fig.  13Cc,  the  bar  is 
brought  into  tension  by  the  load  P  moving  with  the  velocity  V. 
The  effective  work  yPh  is  expended  in  stressing  the  bar  from 
o  up  to  Q  while  the  elongation  increases  from  o  up  to  q,  so  that 
the  stored  stress  energy  at  the  moment  of  greatest  elongation  then 
is  J<2#;  hence  %Qq  =  yPh.  Also  q/e  =  Q/P  if  e  is  the  static 
elongation  due  to  P.  From  these  two  equations  the  values  of 
Q  and  q  for  horizontal  impact  are  found  to  be, 

i          = 


ART.  130 


INERTIA  IN  AXIAL  IMPACT 


333 


in  which  r?  is  the  inertia  coefficient  and  p  is  the  ratio  W/P.  If 
S  is  the  static  unit-stress  P/a  and  T  is  the  dynamic  unit-stress 
Q/a.  then  also  T  =  S(2T)h/e)*.  It  is  seen  that  the  values  of  Q 
and  T  are  less  than  those  given  by  (128)  on  account  of  the  energy 
lost  in  overcoming  inertia.  For  instance  let  T\  be  the  value 
computed  from  the  first  equation  in  (128)",  then  T \-ifi  is  the 
value  when  the  resistance  of  inertia  is  taken  into  account;  thus 
when  W  equals  P,  the  dynamic  unit-stress  is  0.77  J1!,  and  when 
W  is  equal  to  ^P  it  is  0.51  TV 


«»"*"" 

\ 

/ 

I 

/'-v 

-  ^^               II       I  / 
X    N                ~l    ~3 

1 

(     m       Kf 

s-"™—  q—  - 

-*• 

V  ^                1    1        K 

1       1       S 

Fig.  130a  Fig.  1306 


Fig.   130c 


For  the  case  of  vertical  impact  shown  in  Fig.  1306  the  weight 
P  falls  through  the  height  h  upon  the  end  of  the  bar.  Here  the 
equations  are  the  same  as  in  the  last  paragraph  except  that  Pq 
is  to  be  added  to  the  second  member  of  the  energy  equation,  since 
P  falls  through  the  distance  q  after  striking  the  end  of  the  bar, 
whence  \Qq  =  P(f)h-\-q).  Solving  the  equations  there  results, 

(130)' 


as  the  formulas  for  vertical  impact  in  which  the  impact  coefficient 
y  has  the  value  given  above.  As  a  numerical  example  let  it  be 
required  to  find  the  dynamic  stress  produced  in  a  vertical  wrought- 
iron  bar,  two  square  inches  in  section  area  and  18  feet  long,  by  a 
body  weighing  600  pounds  and  falling  through  the  height  of  one 
foot.  Here  1^=10X6X2  =  120  pounds  (Art.  17),  ,0  =  0.2  and 
77=0.882;  also  e  =  Pl/aE  =  0.002  59  inches;  then  the  formula  gives 
(3  =  54840  pounds.  The  stress  due  to  the  static  load  of  600 
pounds  is  S  =  300  pounds  per  square  inch,  but  the  dynamic  stress 
due  to  the  same  load  falling  through  a  height  of  one  foot  is  T  = 
27  420  pounds  per  square  inch  or  about  90  times  as  great. 


334  IMPACT  AND  FATIGUE  CHAP.  XIV 

When  the  load  P  strikes  upon  a  shelf  or  scale-pan  of  weight 
W\  at  the  free  end  of  the  bar,  the  loss  of  kinetic  energy  is  greater 
than  before,  since  the  inertia  of  a  greater  weight  must  be  over- 
come. Then  at  the  moment  of  complete  contact  W\  has  the 
kinetic  energy  W\  .  v2/2g  and  this  must  be  added  to  the  first 
value  of  K  in  this  article;  also  the  equation  of  impact  becomes 
PV=(P  +  Wi  +  %W)v.  By  the  same  reasoning  as  before  the 
same  equations  are  deduced,  except  that  the  inertia  coefficient  has 
the  value, 

_Wi  _W 

PI=P  p-  P 


For  example,  take  the  case  of  the  wrought-iron  bar  in  the  last 
paragraph  and  let  the  load  of  600  pounds  fall  in  a  scale-pan  which 
weighs  300  pounds.  Here  ,01=0.5,  p  =  o.2,  and  77=0.612;  then 
<2  =  45  I0°  pounds,  and  T  =  22  550  pounds  per  square  inch,  so 
that  the  addition  of  the  scale-pan  diminishes  the  dynamic  stress 
about  1  8  percent.  This  is  a  principle  of  importance  in  bridge  con- 
struction; for  example,  a  heavy  floor  for  a  suspension  bridge 
decreases  the  dynamic  stress  which  may  be  brought  by  the  live 
load  upon  the  vertical  rods  that  connect  the  floor  to  the  cable. 

Prob.  130.  A  weight  of  60  pounds  impinges  upon  the  end  of  a  hori- 
zontal bar  of  wrought  iron  which  is  2  inches  in  diameter  and  12  feet 
long.  Find  the  velocity  of  the  weight  which  will  stress  the  bar  up  to 
its  elastic  limit. 


ART.  131.    INERTIA  IN  TRANSVERSE  IMPACT 

When  a  weight  P  strikes  a  beam  with  the  velocity  V,  it  has 
the  kinetic  energy  P  .  V2/2g,  and  part  of  this  is  lost  in  the  impact, 
while  the  remainder  causes  the  beam  to  deflect  the  amount  q 
which  is  greater  than  the  deflection  /  due  to  a  static  load  P.  Let 
h  be  the  height  of  fall  which  will  produce  the  velocity  V,  and  y 
be  the  ineilia  coefficient,  so  that  yPh  is  the  effective  work  which 
deflects  and  stresses  the  bar.  Then  the  formulas  of  Art.  129 
will  apply  if  yh  is  substituted  for  h.  The  value  of/  will  now  be 
found  for  a  simple  beam  in  a  manner  similar  to  that  followed 


ART.  131  INERTIA  IN  TRANSVERSE  IMPACT  335 

for  axial  impact  in  the  last  article,  and  under  the  same  assump- 
tions. 

Let  the  weight  P  strike  the  middle  of  the  simple  beam  with 
the  velocity  F;  when  complete  contact  is  obtained  both  P  and 
the  middle  of  the  beam  are  moving  with  a  velocity  v  which  is 
less  than  F,  and  any  other  point  of  the  beam  is  moving  with  a 
velocity  u.  Let  W  be  the  weight  of  the  beam,  /  its  length,  x 
any  distance  from  the  left  end,  and  dW  the  element  which  is 
moving  with  the  velocity  u ;  then  dW  =  W  .  dx/L  When  the  middle 
point  has  the  velocity  v  and  the  deflection  y\,  the  element  dW 
has  the  deflection  y  and  the  velocity  u  =  v.y/y\.  Then  the 
kinetic  energy  stored  in  load  and  beam  at  the  instant  of  complete 
contact  is, 


2g    J        2g 

Now  to  find  the  value  of  the  integral,  it  is  assumed  that  the  elastic 
curve  has  the  same  equation  under  a  dynamic  as  under  a  static 
load ;  from  Art.  55  the  ordinate  y  of  the  elastic  curve  in  terms 
of  the  abscissa  x  and  the  maximum  deflection  y,  is  found  to 
be  y=y\($l?x  —  4#3)//3.  Then,  extending  the  integration  over 
the  entire  beam, 

-H 

and  accordingly  the  kinetic  energy  which  is  available  for  deflec- 
tion and  stress  is, 


To  find  the  value  of  v  in  terms  of  F,  the  same  reasoning  will 
be  followed  as  in  the  last  article.  Instead  of  PV  =  (P  +  Pi)v 
for  the  impact  of  P  against  a  free  body  PI  at  rest,  there  must  be 
written  for  the  beam, 


PV=Pv+   fdW  .  u=Pv+?^f  (3Px 

and  hence  the  velocity  at  the  instant  of  complete  contact  of  beam 
and  load  is  v  =  V/(i+  f  W/P).   Inserting  this  in  the  above  expres- 


336  IMPACT  AND  FATIGUE  CHAP.  XIV 

sion  for  the  effective  work,  there  is  found, 

- 

in  which  T?  is  the  inertia  coefficient  and  p  is  the  ratio  TF/P. 

The  case  of  horizontal  impact  against  a  beam  occurs  in  some 
testing  machines,  the  ends  of  the  bar  being  prevented  from  mov- 
ing sidewise  while  a  hammer  strikes  horizontally  against  the 
middle  of  the  beam.  Here  the  discussion  is  the  same  as  that  in 
Art.  129  except  that  h  is  replaced  by  yh,  and 

T  =  S(27)h/f)*  q=(2T)hf)*  (131) 

are  the  formulas  for  dynamic  flexural  unit-stress  and  dynamic 
deflection.  Here  S  is  the  static  stress  found  from  the  flexure 
formula  (41)  and  /  is  the  static  deflection  found  from  Art.  55 
for  the  load  P,  and  T?  is  the  impact  coefficient  whose  value  is 
given  above. 

For  the  case  of  vertical  impact  where  the  load  P  falls  through 
the  height  h  upon  the  middle  of  the  beam,  the  discussion  is  also 
the  same  as  that  in  Art.  129,  except  that  h  is  replaced  by  yh,  and 

T  =  S+S(i  +  27?  V/)*  <?=/+  (/2+  2^/)*  (131)' 

are  the  formulas  for  vertical  impact,  in  which  5,  /,  and  T?  have 
the  same  signification  as  before.  All  the  formulas  of  this  article 
are  only  valid  when  the  unit-stress  T  is  less  than  the  elastic  limit 
of  the  material. 

As  a  numerical  example,  let  a  cast-iron  simple  beam  of  36 
inches  span  and  2X2  inches  in  section  have  a  load  of  50  pounds 
at  the  middle.  The  flexural  unit-stress  and  the  deflection  due 
to  this  static  load  are  found  from  Arts.  48  and  55  to  be, 

5  =  338  pounds  per  square  inch,  /=  0.00243  inches 

Now  let  the  ends  of  the  beam  be  prevented  from  moving  side- 
wise  when  a  horizontally  moving  load  of  50  pounds  strikes  it 
at  the  middle  with  a  velocity  due  to  a  fall  of  2  inches.  Then, 
disregarding  the  inertia  of  the  beam,  formulas  (129)'  give  the 
dynamic  stress  and  deflection, 

r=i3  700  pounds  per  square  inch,  5  =  0.0986  inches 


ART.  131  INERTIA  IN  TRANSVERSE  IMPACT  337 

Taking  the  inertia  of  the  beam  into  account,  the  weight  W  is 
37.6  pounds,  <o=  1^/^  =  0.752,  9  =  0.632,  and  then  formulas  (131) 
give, 

T=  10  900  pounds  per  square  inch,  #=0.0784  inches 

These  values  of  T  are  greater  than  the  elastic  limit  of  cast  iron 
and  hence  cannot  be  relied  upon  as  exact.  The  example  shows, 
however,  that  small  velocities  of  impact  may  produce  high  dy- 
namic stresses  in  a  beam. 

When  the  load  P  falls  into  a  scale  -pan  of  weight  Wi  which 
is  attached  to  the  middle  of  the  simple  beam,  the  loss  in  impact 
is  less  than  when  it  falls  directly  upon  the  beam.  For  this  case 
the  above  reasoning  is  to  be  modified  by  replacing  P  by  P  +  Wi 
in  the  first  value  of  K,  and  also  in  the  second  member  of  PV  = 
(P+Pi)v.  The  ineiuia  coefficient  will  then  be  given  by, 


_  _ 

pl~  P      p~~p 

For  example,  taking  the  cast-iron  beam  of  the  last  paragraph, 
let  the  weight  of  50  pounds  fall  vertically  upon  its  middle  from 
a  height  of  2  inches,  there  being  no  scale-pan;  then  77  =  0.632 
and  from  (131)', 

7^=11  240  pounds  per  square  inch,  #  =  0.0808  inches 

Now  let  there  be  a  scale-pan  weighing  20  pounds  into  which  the 
load  of  50  pounds  falls  from  a  height  of  2  inches;  here  ^i  =0.400? 
p  =  0.752,  77  =  0.505,  and  then  (131)'  give, 

T=  10  090  pounds  per  square  inch,  #=0.0725  inches 

which  show  that  the  effect  of  the  scale-pan  is  materially  to  diminish 
the  dynamic  stress  due  to  impact. 

The  numbers  $-J  and  f  in  the  inertia  coefficient  apply  only 
to  a  simple  beam  struck  at  the  middle  :  they  do  not  apply,  how- 
ever, to  other  points  than  the  middle  of  the  span.  For  a  beam 
with  fixed  ends  impinged  upon  at  the  middle,  ||  is  to  be  used 
instead  of  £}  and  J  instead  of  f.  For  a  cantilever  beam  struck 
at  the  end  by  a  load,  trV  takes  the  place  of  £f  and  f  that  of  J. 

Prob.  1310.  Verify  the  statements  in  the  preceding  sentence. 


338  IMPACT  AND  FATIGUE  CHAP,  xiv 

Prob.  1316.  Compute  the  dynamic  deflection  for  Prob.  1296,  tak- 
ing into  account  the  inertia  of  the  beam. 

ART.  132.    VIBRATIONS  AFTER  IMPACT 

Referring  to  the  case  of  axial  impact  shown  in  Fig.  130c,  it 
is  clear  that  the  velocity  of  the  end  of  the  bar  at  any  instant  is  a 
function  of  the  elongation  x.  When  x  equals  the  final  elonga- 
tion q,  the  velocity  is  zero;  the  end  of  the  bar  then  springs  back 
and  increases  until  x  is  zero,  and  then  decreases  until  it  becomes 
zero  when  x  is  equal  to  —  <?;  the  vibration  is  next  performed 
in  the  opposite  direction.  These  vibrations  would  continue 
indefinitely  were  it  not  for  air  resistance  and  molecular  friction, 
but  owing  to  such  resistances  they  become  less  and  less  in  ampli- 
tude, until  finally  the  bar  comes  to  rest.  Neglecting  the  weight 
of  the  bar  in  comparison  with  the  load,  the  following  investi- 
gation gives  the  time  of  one  vibration. 

Let  vx  be  the  velocity  of  the  end  of  the  bar  when  the  elonga- 
tion x  is  attained,  and  let  Qx  be  the  corresponding  stress  in  the 
bar.  The  kinetic  energy  of  the  moving  weight  then  equals  the 
internal  work  still  to  be  stored  in  the  bar  in  increasing  oc  to  q, 
or  P  .  vx2/2g  =  $Qq—%QAx.  Replacing  Q  by  P  .  q/e  and  Qx  by 
P  .  x/e,  where  e  is  the  static  elongation  due  to  P,  this  equa- 
tion becomes, 

v*2  =  (?-*?)g/e  or  (dx/dt)*  =  (f-x?)g/e 

which  gives  the  velocity  of  the  end  of  the  bar  for  any  value  of  x\ 
this  velocity  is  zero  when  x  =  +q  and  also  when  x=  —  q.  To 
find  the  time  in  which  this  vibration  is  performed,  let  /  be  the 
number  of  seconds  counted  from  the  instant  when  x  =  q;  the 
velocity  at  any  other  instant  is  then  dx/dt,  as  already  indicated 
in  the  last  equation,  which  may  be  written, 

dt     f    e/g  \*  (e\*  x 

T-  =  l  9       9}  or  /=-    arcsm- 

dx     \q2—x2J  \gj  q 

and  taking  the  integral  between  the  limits  +  q  and  —  q  there 
is  found  /  =  n(e/g)*  as  the  time  of  one  vibration.  This  is  the 
time  of  one  vibration  of  a  pendulum  which  has  the  length  e. 


ART.  132  VIBRATIONS    AFTER   IMPACT  339 

This  time  is  the  same  for  all  subsequent  vibrations  notwith- 
standing that  the  amplitude  q  becomes  less  and  less  on  account 
of  the  air  resistance. 

For  the  case  of  the  vertical  bar,  the  first  equation  of  this  article 
will  be  modified  by  adding  P(q—x)  to  the  first  member,  this 
expressing  the  work  still  to  be  performed  while  the  load  P  falls 
through  the  distance  q— x.  Then  are  found, 

vx2=^ (q2-x2)-2g(q-x)  /=(-j*arc  sin^+C 

the  second  being  derived  from  the  first  by  replacing  vx  by  dx/dt 
and  integrating  as  above.  By  discussing  the  first  equation,  it 
is  found  that  vx  is  a  maximum  when  x  =  e  as  indicated  in  Fig.  130&, 
and  that  vx  is  zero  when  x  =  q.  Hence,  counting  the  time  from 
the  instant  when  x  =  c,  the  time  elapsed  when  x  =  q  is  %n(e/g)*y 
which  is  that  of  one-half  a  vibration.  In  the  backward  vibration 
the  end  of  the  bar  moves  from  x  =  q  to  x  =  e  in  the  same  time; 
the  first  equation  shows,  however,  that  v=o  when  x=—  q  +  2e, 
so  that  the  end  of  the  bar  moves  upward  and  the  time  elapsed 
between  x  =  q  and  x=—  q  +  2e  is  n(e/g)*,  which  is  the  time  of 
one  vibration.  The  end  of  the  bar  hence  oscillates  back  and 
forth  about  the  point  for  which  x  =  e,  that  is,  about  the  point 
where  it  finally  comes  to  rest;  while  the  amplitude  q  —  e  of  the 
vibrations  grows  less  and  less,  the  time  of  each  vibration  remains 
the  same,  namely,  that  of  a  simple  pendulum  having  a  length 
e  equal  to  the  elongation  due  to  the  static  load  P. 

The  above  conclusion  regarding  the  time  of  vibration  will 
be  slightly  modified  when  the  inertia  of  the  bar  is  taken  into 
account.  Let  W  be  the  weight  of  the  bar  and  W\  that  of  the  scale- 
pan  at  its  end.  Then  when  P  and  W\  have  the  velocity  vx  the 
kinetic  energy  of  the  moving  particles  is  (P  +  Wi+$W)vx2/2g; 
for  the  vertical  bar  P(q  —  x)  is  to  be  added  to  this  to  give  the  total 
work  still  to  be  performed.  This  sum  is  to  be  equated  to  the  stress 
energy  %Qq  —  \Qxx  which  is  to  be  stored  in  the  bar  in  increasing 
the  elongation  from  x  to  q.  Stating  this  equation,  there  is  found, 


340  IMPACT  AND  FATIGUE  CHAP,  xiv 

and  by  the  same  method  as  before  there  results, 


as  the  time  of  one  vibration,  in  which  e'  is  the  static  elongation 
due  to  P  +  Wi+^W.  This  formula  also  applies  to  the  horizontal 
bar.  As  a  numerical  example,  let  the  data  of  Problem  130  be 
considered.  Here  P=6o  pounds,  ^  =  125.7  pounds,  TFi=o, 
and  the  static  elongation  due  to  P  +  ^W  is  e'  =0.000  187  inch. 
Then,  taking  g  as  32.  2X12  inches  per  second  per  second,  the  time 
of  one  vibration  is  /=  0.002  2  seconds,  and  about  460  oscillations 
would  be  performed  in  one  second  if  there  were  no  frictional 
resistances. 

The  vibrations  of  a  beam  after  the  impact  of  a  load  are  In 
all  respects  similar  to  those  of  a  bar.  By  investigations  exactly 
like  those  for  the  bar,  it  may  be  shown  that  the  time  of  one  vibra- 
tion for  a  simple  beam  struck  at  the  middle  is, 


where  /is  the  static  deflection  due  to  P;  here  /9/is  the  static  de- 
flection due  to  P  +  Wi+^lW.  When  the  load  remains  on  the 
beam  after  impact,  the  vibrations  occur  about  the  position  which 
it  finally  assumes  under  the  static  load  ;  when  it  does  not  remain 
on  the  beam,  the  vibrations  occur  about  the  position  that  it  had 
before  the  impact.  Fig.  133  shows  the  vibrations  of  a  railroad 
rail  after  impact. 

The  above  formulas  do  not  apply  to  the  incomplete  semi- 
vibration  which  occurs  during  the  impact  while  the  end  of  the 
bar  is  descending  through  the  distance  e  or  the  middle  of  the 
beam  through  the  distance  /.  The  time  of  all  subsequent  vibra- 
tions is  the  same  whatever  be  their  amplitude  and  is  equal  to  that 
of  a  simple  pendulum  which  has  the  length  fie  for  the  bar  or  ftf 
for  the  beam.  The  shorter  this  length  the  less  is  the  time  /  and 
the  more  rapid  are  the  vibrations. 

Prob.  132.  When  a  load  falls  upon  a  beam  show  that  the  amplitude 
of  the  first  complete  vibration  is  a  little  less  than  2  (/2+  2  >?/*/)*. 


.  133  EXPERIMENTS  ON  ELASTIC  IMPACT  341 


ART.  133.    EXPERIMENTS  ON  ELASTIC  IMPACT 

Numerous  experiments  have  been  made  to  test  the  formulas 
for  elastic  impact  derived  in  the  preceding  articles,  and  a  dis- 
cussion of  some  of  them  will  now  be  given.  These  formulas 
are  not  exact,  because  it  has  been  supposed  in  their  deduction 
that  the  velocity  of  transmission  of  stress  is  infinite,  and  that  no 
loss  of  energy  occurs  except  in  impact.  The  assumption  regard- 
ing velocity  of  stress  does  not  lead  to  any  appreciable  error,  but 
that  regarding  loss  of  energy  may  do  so  in  cases  where  the  falling 
body  is  deformed  so  that  it  absorbs  energy  or  where  a  portion  of 
the  energy  is  expended  in  deforming  the  supports  of  the  bar  or 
beam. 

Simple  experiments  may  be  made  by  the  student,  using  a 
common  spiral  spring  instead  of  a  bar,  so  that  the  elongations 
may  be  easily  measured.  For  example,  a  spiral  spring  about 
32  inches  long  and  weighing  0.6  ounces  was  found  to  elongate 
0.390  inches  under  a  static  load  of  10  ounces.  When  loaded  with 
8  ounces  and  the  end  depressed  by  the  hand  and  then  released, 
there  were  counted  304  vibrations  in  100  seconds,  when  loaded 
with  14  ounces  there  occurred  230  vibrations  in  100  seconds. 
Here  the  actual  times  of  one  vibration  were  0.329  and  0.435 
second,  while  formula  (132)  gives  0.321  and  0.428  seconds,  so 
that  the  agreement  of  experience  and  theory  is  very  fair. 

The  simplest  case  of  impact  on  beams  is  that  of  a  single 
sudden  load  which  was  discussed  in  Art.  127.  Kirkaldy  made 
experiments,  about  1860,  to  test  the  theoretic  law  that  the 
deflection  under  such  a  load  is  double  that  due  to  an  equal 
static  load.  A  load  was  attached  to  a  ring  placed  around  the 
middle  of  the  beam  and  the  ring  supported  so  that  its  lower 
surface  just  touched  the  upper  surface  of  the  beam;  the  sup- 
port of  the  ring  was  then  suddenly  withdrawn  so  that  the  load 
acted  with  its  full  intensity  during  the  entire  period  of  the  deflec- 
tion, which  was  registered  upon  a  vertical  sheet  of  paper  by  a 
pencil  screwed  to  the  side  of  the  beam.  Before  applying  the 
loads  in  this  sudden  manner,  the  deflections  due  to  gradually 


Size  of  Beam 

Load 

Deflection  in 

Inches 

Inches 

Pounds 

Gradual 

Sudden 

1X2 

112 

0-255 

0.515 

1X2 

224 

0.580 

0-933 

lX3 

224 

0.163 

0.303 

lX3 

560 

0.410 

0.720 

4XlJ 

448 

0.770 

1.510 

4Xii 

784 

I-275 

2.225 

342  IMPACT  AND  FATIGUE  CHAP.  XIV 

applied  loads  were  measured.  The  beams  were  of  cast  iron  and 
laid  on  supports  9  feet  apart.  The  results  here  shown  are  the 
mean  of  two  or  three  tests  upon  different  beams.  It  will  be 
found  that  for  each  size  of  beam  the  first  load  gives  a  unit-stress 
less  than  the  elastic  limit,  while  the  second  gives  a  greater  value. 

Ratio  of  Deflections 
Sudden  to  Gradual 

2.04 

1.61 

1.86 
1.76 
1.96 
i-73 

Thus,  for  the  first  beam  under  112  pounds  at  the  middle,  the 
unit-stress  S  computed  from  the  flexure  formula  is  2270,  while 
for  224  pounds  it  is  4540  pounds  per  square  inch.  The  aver- 
age of  the  three  ratios  of  the  deflections  for  the  beams  in  which 
the  elastic  limit  was  not  exceeded  is  1.95,  which  is  a  fair  agree- 
ment with  the  theoretic  number  2. 

It  was  early  recognized  that  the  inertia  of  a  bar  or  beam 
diminished  the  theoretic  stress  and  deflection.  About  1830  the 
inertia  coefficient  y  was  thought  to  be  of  the  form  >?  = 
i/(i+  p  -  W/P)  and  Hodgkinson  determined  by  experiments  on 
beams  that  the  value  of  /*  was  approximately  J,  although  the 
theoretic  number  ^  J  was  not  derived  until  several  years  later  by 
Cox.  This  form  of  the  inertia  coefficient  has  been  generally 
used  since,  and  it  was  employed  in  previous  editions  of  this  book. 
The  form  deduced  in  this  edition  is  believed  to  be  more  exact 
according  to  the  theory  of  impact. 

The  experiments  made  by  Keep  in  1899  on  tool- steel  bars 
under  both  horizontal  and  vertical  impact  enable  an  interesting 
comparison  of  theory  and  practice  to  be  made.  In  one  series 
of  tests  bars  1X1X24  inches  were  loaded  with  weights  of  25, 
50,  75,  and  100  pounds  and  the  corresponding  static  deflections 
were  found  to  be  0.0028,  0.0056,  0.0084,  and  0.0112  inches.  They 
were  then  struck  laterally  by  hammers  of  the  same  weights  which 


ART.  133  EXPERIMENTS   ON    ELASTIC    IMPACT  343 

swung  like  pendulums  and  had  a  vertical  fall  of  2  inches.  The 
dynamic  deflections  due  to  these  weights,  as  carefully  measured 
by  a  graphic  recording  apparatus,  are  given  below.  From  the 
given  data  and  observed  static  deflections  the  deflections  under 
impact  have  been  computed  from  formula  (131)",  and  the  follow- 
ing is  a  comparison  of  observed  and  computed  values : 

Swinging    P=    25  50  75  100   pounds 

Observed     (7=0.122         0.150        0.175         0.200  inches 
Computed  q  =  o.ogj         0.142         0.178         0.207  inches 

Experiments  were  also  made  by  allowing  the  same  weights  to 
fall  vertically  on  the  bars  through  heights  of  2  inches.  For- 
mula (131)'  applies  to  this  case,  and  the  following  is  a  com- 
parison of  the  dynamic  deflections  as  observed  and  computed; 

Falling       ^=25  50  75  100  pounds 

Observed     <?= 0.130        0.159         0.181         0.209  inches 
Computed  </  =  o.ioo        0.150        0.186        0.219  inches 

It  is  seen  that  the  comparison  is  very  satisfactory  except  for  the 
lightest  hammer;  it  would  be  expected,  however,  that  the  observed 
values  should  be  always  slightly  less  than  the  theoretic  ones, 
because  some  of  the  work  of  a  falling  weight  is  probably  expended 
in  deforming  and  heating  that  weight. 

Fig.  133  gives  an  autographic  record  of  the  deflection  of  a 
railroad  rail  and  its  subsequent  vibrations,  taken  by  P.  H.  Dud- 
ley in  1895.  A  rail  30  feet  long  and  weighing  80  pounds  per 
yard  was  placed  at  its  extreme  ends  upon  rigid  supports 
and  had  a  scale-pan  weighing  210  pounds  hung  from  the 
middle.  Secured  to  the  rail  was  an  attachment  carrying  a 
pencil,  which  recorded  the  deflection  and  vibrations  upon  hori- 
zontally moving  paper.  A  load  of  100  pounds  being  suddenly 
applied,  the  dynamic  deflection  was  0.24  inches,  but  when  the 
beam  came  to  rest  the  static  deflection  was  0.12  inches.  The 
weight  of  100  pounds  was  then  dropped  upon  the  scale-pan 
from  a  height  of  12  inches,  and  the  maximum  deflection  was 
0.91  inches;  about  240  vibrations  then  ensued  and  in  about 
42  seconds  the  rail  came  to  rest.  Applying  formula  (131)'  and 
to  this  case,  there  are  found  p  =  8,  /PI  =2.1,  TJ  =0.1067,  and 


344 


IMPACT  AND  FATIGUE 


CHAP.  XIV 


q=o.6g  inches  for  the   dynamic  deflection,  which  is  not  a  good 
agreement  with  the  observed  value.     Applying  formula  (132)  there 

results  /  =  0.146  for  the  time  of  one 
vibration,  and  hence  the  theoretic  time 
for  240  observations  is  about  35 
seconds.  While  the  numerical  results 
derived  from  theory  do  not  agree  very 
well  with  the  observations,  this  experi- 
ment is  a  very  instructive  one,  and 
the  figure  shows  how  a  beam  vibrates 
back  and  forth  about  the  position 
that  it  occupies  after  coming  to  rest. 

Prob.  133.  Explain  all  the  lines  and 
notes  on  the  left-hand  part  of  Fig.  133. 
See  Railroad  Gazette,  May  31,  1905. 

ART.  134.  PRESSURE  DURING  IMPACT 

When  a  weight  P  falls  from  a 
height  h  upon  the  end  of  a  vertical 
bar,  a  pressure  is  produced  which  is 
equal  at  any  instant  to  the  stress  then 
existing  in  the  bar.  For  the  case  of 
elastic  elongation  discussed  in  Art. 
130,  the  maximum  stress  is  (J,  which 
occurs  when  the  greatest  elongation  q 
is  attained.  The  stress  Qx  which 
occurs  for  any  elongation  x  is  equal 
to  P  .  x/e,  where  e  is  the  static  elon- 
gation due  to  P.  Thus  the  pressure 
increases  directly  with  x,  becomes  P 
when  x  =  ey  and  reaches  its  maximum 
value  Q  when  the  greatest  elongation 
q  is  reached.  Similarly,  Art.  131  shows  that  for  a  beam  under 
elastic  impact,  the  pressure  on  the  beam  increases  directly  as  the 
deflection  y,  becomes  equal  to  P  when  y=J\  and  reaches  its 
maximum  value  when  the  maximum  deflection  q  is  attained. 


A.RT.  134  PRESSURE  DURING  IMPACT  345 

The  actual  forces  acting  between  the  falling  load  and  the  end 
of  the  bar  differ  somewhat  from  the  stress  in  the  bar,  because 
there  exists  a  pressure  which  overcomes  inertia  during  the  first 
part  of  the  fall.  The  exact  determination  of  the  actual  pressure 
for  the  case  of  elastic  impact  is  a  problem  of  so  much  complexity 
that  it  will  not  be  undertaken  here,  while  its  determination  is 
impossible  by  theory  for  cases  where  the  elastic  limit  of  the  mate- 
rial is  exceeded.  From  the  point  of  view  of  the  engineer,  the 
pressures  that  cause  motion  of  the  bar  or  beam  are  of  little  im- 
portance compared  with  those  that  cause  deformation  of  the 
material. 

The  following  method  is  sometimes  used  for  estimating  the 
mean  pressure  on  a  beam  during  its  deflection  under  impact. 
Let  R  be  this  mean  or  average  pressure  which  is  exerted  through 
the  deflection  q.  Then  Rq  is  the  work  performed  by  it,  while 
that  done  by  the  falling  load  is  P(h  +  q).  Placing  these  equal 
there  results  R  =  P(i  +  h/q)  for  the  mean  pressure.  While  this 
expression  is  correct  for  both  elastic  and  non-elastic  deflections,  it 
must  be  borne  in  mind  that  it  does  not  give  the  mean  compressive 
stress  on  the  upper  surface  of  the  beam  where  the  impact  occurs, 
but  always  a  greater  value,  because  some  of  the  pressure  is  ex- 
erted in  overcoming  inertia. 

To  illustrate,  take  the  case  of  the  railroad  rail  discussed  at  the 
end  of  the  last  article,  where  a  weight  of  100  pounds  fell  from  a 
height  of  12  inches  and  caused  an  elastic  deflection  of  0.91  inches. 
Then  R=  looX  12.91/0.91  =  i  420  pounds  for  the  average  pressure 
during  the  dynamic  deflection.  The  static  load  Q  which  will 
produce  the  same  deflection  is  (3  =  100X0.91/0.12  =  760  pounds, 
so  that  the  average  pressure  which  was  effective  in  stressing  the 
beam  and  causing  the  deflection  was  380  pounds.  Undoubtedly, 
the  actual  average  pressure  was  about  i  420  pounds,  but  more 
than  two -thirds  of  this  was  expended  in  overcoming  the  inertia 
of  the  heavy  beam. 

In  all  cases  of  elastic  impact,  the  mean  or  average  stress  is 
one-half  of  the  maximum,  because  the  stress  increases  uniformly 
with  the  deflection.  This  is  not  true  when  the  elastic  limit  of 


346  IMPACT  AND  FATIGUE  CHAP,  xiv 

the  material  is  exceeded,  and  in  general  the  mean  stress  is  less 
than  one-half  of  the  maximum.  The  same  probably  holds  for 
the  total  pressure  that  is  exerted  both  to  overcome  inertia  and 
to  cause  stress.  The  maximum  unit-pressure  that  acts  between 
the  surfaces  of  contact  will  depend,  of  course,  upon  the  area  of 
contact  and  upon  the  manner  in  which  the  total  pressure  is  dis- 
tributed over  that  area. 

Prob.  1340.  A  vertical  bar  weighing  27  pounds  receives  a  stress  of 
196  pounds  when  a  load  of  100  pounds  acts  axially  upon  it.  Through 
what  height  must  a  load  of  50  pounds  drop  in  order  to  produce  the 
same  stress? 

Prob.  1346.  A  ram  weighing  2  ooo  pounds  falls  from  a  height  of  20 
feet  upon  a  railroad  rail  laid  on  supports  6  feet  apart,  this  being  one 
method  of  testing  rails  at  the  mill.  Compute  the  average  pressure 
when  the  ram  deflects  a  heavy  rail  i\  inches;  also  when  it  deflects  a 
lighter  rail  5  inches. 


ART.  135.    IMPACT  CAUSING  RUPTURE 

The  cases  of  impact  thus  far  considered  have  been  those  where 
the  greatest  unit-stress  does  not  exceed  the  elastic  limit  of  the 
material.  It  is,  however,  easy  to  cause  the  rupture  of  a  bar  or 
beam  by  allowing  a  heavy  load  to  drop  upon  it  from  a  sufficient 
height.  For  such  cases  theory  furnishes  no  formulas  and  experi- 
ment is  the  only  source  of  information.  Many  tests  have  been 
made  to  ascertain  the  phenomena  of  rupture  under  impact,  and 
the  general  conclusions  derived  will  now  be  stated. 

In  1807  Thomas  Young  announced  the  fundamental  ideas  of 
the  resistance  of  materials  under  impact.  "The  action  which 
resists  pressure,"  he  said,  "is  called  strength,  and  that  which 
resists  impulse  may  properly  be  called  resilience."  He  stated 
that  the  resilience  of  a  body  is  proportional  to  its  strength  and 
extension  jointly,  and  that  it  is  measured  by  the  height  through 
which  a  given  weight  must  fall  to  cause  rupture.  The  resili- 
ence of  beams  of  the  same  kind  he  made  proportional  to  their 
volumes,  as  also  the  resilience  of  shafts,  whether  solid  or  hollow. 


ART.  135  IMPACT   CAUSING    RUPTURE  347 

At  that  time  the  elastic  limit  of  materials  was  only  vaguely  recog- 
nized, and  Hooke's  law  of  proportionality  of  deformation  to  stress 
was  often  applied  to  all  the  phenomena  preceding  rupture.  Young's 
statements  are  valid  in  a  general  way,  but  it  is  now  known  that  there 
are  two  divisions  of  the  subject  of  impact:  first,  that  where  the 
elastic  limit  is  not  exceeded  and  where  the  term  resilience  is  properly 
applicable,  and  second,  that  where  the  elastic  limit  is  exceeded 
and  rupture  finally  occurs. 

In  1818  experiments  were  made  by  Tredgold  on  wooden 
beams  subject  to  the  impact  of  a  falling  ball,  and  he  concluded 
that  the  work  required  to  produce  rupture  was  not  proportional 
to  the  volume  of  the  bar.  Hodgkinson  in  1835  experimented  on 
cast-iron  beams  and  found  that  the  deflections  seemed  to  be 
proportional  to  the  velocities  of  the  falling  weight.  In  1849 
were  published  the  results  of  an  extensive  series  of  experiments 
made  by  a  British  commission,  and  here  the  influence  of  inertia 
in  diminishing  the  deflection  of  a  beam  under  impact  was  fully 
recognized.  Kirkaldy  in  1862  made  experiments  on  axial  impact 
by  sudden  loads,  and  found  that  some  bars  were  broken  under 
loads  less  than  those  required  when  slowly  applied.  The  impact 
hammer  or  ram,  introduced  by  Sandberg  and  Styffe  in  1868  for 
testing  railroad  rails,  has  proved  valuable  for  comparative  pur- 
poses since  the  information  obtained  is  similar  to  that  derived 
from  the  cold-bend  test.  Maitland,  in  1887,  showed  by  many 
experiments  on  tensile  specimens  subject  to  many  blows  of  a  falling 
ram,  that  the  ultimate  elongation  was  much  greater  than  in  static 
tests;  the  use  of  many  blows  to  cause  rupture  introduces,  how- 
ever, complications,  and  it  has  been  found  that  the  best  plan  to 
obtain  valid  results  is  to  use  a  load  and  fall  which  is  just  suffi- 
cient to  break  the  specimen  at  one  blow. 

When  a  specimen  is  broken  under  tensile  impact,  the  work 
expended  may  be  ascertained  by  measuring  the  area  of  a  stress 
diagram  which  is  autographically  drawn  by  the  machines  and 
which  also  shows  the  ultimate  elongation.  Experiments  of  this 
kind  made  by  Hatt  in  1904  on  various  kinds  of  s'eel  have  shown 
that  the  work  required  to  rupture  a  bar  by  impact  is  usually 


348  IMPACT  AND  FATIGUE  CHAP.  XIV 

greater  than  that  in  common  static  tests  where  the  load  is  gradu- 
ally applied.  From  the  mean  of  about  170  tests,  Hatt  found 
that  the  average  work  required  for  rupture  by  impact  was  30  per- 
cent greater  than  in  static  tests,  and  that  the  ultimate  elonga- 
tion under  rupture  was  20  percent  greater.  The  fracture  was 
similar  in  both  impact  and  static  tests,  but  in  the  former  there 
were  often  observed  two  or  more  places  of  marked  diminished 
section,  whereas  only  one  occurred  in  the  latter.  For  round 
bars  of  soft  steel  there  appeared  to  be  little  difference  in  ulti- 
mate elongation  and  work  whether  the  bars  were  broken  in  ten 
minutes  by  the  common  method  or  in  one  one-hundredth  of  a 
second  by  impact. 

When  a  body  is  ruptured  by  impact,  it  is  important  that  the 
apparatus  should  be  so  arranged  that  all  the  work  of  the  falling 
ram  may  be  expended  on  the  specimen,  and  not  be  absorbed  by 
other  parts  of  the  apparatus.  If  the  weight  falls  upon  a  shelf 
or  pan  connected  to  the  bar  or  beam,  this  should  be  made  heavy 
so  that  work  may  not  be  expended  in  deforming  it.  When  the 
ends  of  a  tensile  specimen  are  larger  than  the  main  part,  they 
should  be  made  very  large  so  as  not  to  absorb  energy,  and  the 
supports  of  a  beam  should  be  made  heavy  for  a  similar  reason. 
Impact  tests  are  of  much  value  in  determining  the  qualify  of 
materials,  and  they  are  widely  used  for  railroad  rails,  car  and 
locomotive  axles,  and  other  steel  pieces  which  are  subject  to 
shocks.  The  impact  tests  introduced  by  Keep  for  cast  iron 
undoubtedly  give  valuable  information  regarding  its  behavior 
under  shocks.  In  general  it  is  probable  that  impact  tests  show 
lack  of  homogeneity  of  the  material  better  than  static  tests. 

Autographic  records  taken  during  a  tensile  impact  test  give 
valuable  information  regarding  the  elastic  limit  and  ultimate 
strength  of  the  material.  The  elastic  limit  is  often  found  to  be 
higher  than  under  static  tests,  while  the  ultimate  strength  is 
usually  a  little  lower,  the  difference  between  these  unit-stresses 
being  much  less  than  in  the  usual  method  of  testing.  For  timber 
Hatt  has  found  that  the  elastic  limit  is  nearly  doubled  undei 
impact. 


ART.  136  STRESSES   DUE   TO  LlVE   LOADS  349 

The  formulas  for  dynamic  stress  and  elongation  deduced  in 
the  preceding  articles  do  not  apply  to  cases  where  the  elastic 
limit  is  exceeded,  and  hence  all  attempts  to  verify  these  formulas 
by  experiments  on  the  rupture  of  bars  and  beams  are  fallacious. 
Hodgkinson  broke  cast-iron  beams  under  both  sudden  and  gradual 
loads,  and  found  that  the  ratio  of  the  latter  to  the  former  was 
always  less  than  2,  which  should  be  the  true  ratio  if  the  elastic 
law  were  applicable.  The  following,  for  example,  are  three 
of  his  results  for  cast-iron  beams  with  a  span  of  9  feet: 

Size  of  Beam  Breaking  Load  in  Pounds  Ratio  of  Gradual 

Inches  Gradual  Sudden  to  Sudden  Load 

1X2  1000  569  1.76 

1X3  2  008  I  219  1.65 

4Xii  I  911  I  082  1.77 

The  discrepancy  between  the  theoretic  ratio  and  the  actual  ones 
is  here  not  very  great  because  the  elongation  for  cast-iron  is 
small  (Fig.  4).  For  wrought  iron,  where  the  elongation  is  large 
and  the  stress  curve  greatly  deviates  from  a  straight  line,  a  greater 
discrepancy  would  be  expected,  and  Kirkaldy  found  that  the 
ratio  of  gradual  to  sudden  load  which  caused  the  rupture  of 
wrought-iron  beams  ranged  from  1.2  to  1.3. 

Prob.  135.  Consult  Hart's  paper  in  Proceedings  of  the  American 
Society  for  Testing  Materials,  Vol.  IV,  1904,  and  describe  the  machine 
used  for  making  tensile  impact  tests. 


ART.  136.     STRESSES  DUE  TO  LIVE  LOADS 

A  beam  or  bridge  is  subject  to  the  action  of  both  dead  and 
live  loads,  the  former  including  its  own  weight  and  the  latter 
the  weight  of  the  people,  vehicles,  or  trains  that  pass  over  it. 
The  flexural  stresses  in  the  beam  are  found  by  the  application  of 
formula  (41)  and  those  in  the  members  of  a  bridge  truss  by  the 
methods  set  forth  in  Roofs  and  Bridges,  Part  I.  These  stresses 
are  usually  computed  for  dead  and  live  loads  separately,  regard- 
ing each  as  a  static  load.  The  live  load,  however,  really  produces 
greater  stresses  than  the  computed  ones  because  it  is  applied 


350  IMPACT  AND  FATIGUE  CHAP.  XIV 

quickly,  and  hence  it  is  customary  to  multiply  the  computed 
static  stresses  by  a  number  called  the  "  coefficient  of  impact  " 
in  order  to  obtain  the  increased  stress  due  to  suddenness  of 
application.  Thus,  let  S  be  any  computed  stress  due  to  the 
given  live  load,  this  being  either  a  unit-stress  or  the  total  stress 
in  a  member;  then  iS  is  the  stress  due  to  the  quickness  with 
which  the  load  is  applied,  and  i  is  the  coefficient  of  impact, 
so  that  the  total  stress  due  to  the  live  load  is  S  +  iS  or  (i  +  z)S. 
This  use  of  the  word  impact  does  not  agree  with  that  of  the 
preceding  articles,  but  it  is  customary  in  bridge  literature; 
really  this  coefficient  of  impact  includes  the  effect  of  lateral  and 
vertical  oscillations  due  to  irregularities  of  the  track  as  well  as 
the  effect  of  quickness  of  application  of  the  live  load. 

Various  methods  are  in  use  for  assigning  values  of  the  coeffi- 
cient i,  but  in  all  of  them  no  attention  is  paid  to  the  time  in 
which  the  stress  is  generated,  and  in  fact  they  rest  upon  no  theo- 
retical basis  except  the  law  that  a  suddenly  applied  load  pro- 
duces double  the  stress  of  a  static  one.  Some  engineers  regard  * 
as  unity  for  all  cases  of  live  load,  and  hence  double  the  stress- 
due  to  the  live  load  in  designing  the  section  areas  of  the  mem- 
bers. Many  others  take  *  as  less  than  unity,  using  higher  values 
for  light  bridges  than  for  heavy  ones,  while  some  make  *  depend 
upon  the  length  of  span  and  take  it  higher  for  short  spans  than 
for  long  ones.  Empirical  formulas  for  z  are  given  in  Roofs 
and  Bridges,  Part  I  (New  York,  1905). 

In  this  important  matter  experience  is  in  advance  of  theory 
since  no  formula  has  yet  been  established  for  the  case  of  a  load  P 
applied  to  a  bar  in  a  given  time.  When  slowly  applied  P  pro- 
duces a  unit-stress  5,  when  suddenly  applied  it  produces  a  unit- 
stress  28,  if  the  elastic  limit  is  not  exceeded  (Art.  127);  when 
applied  in  a  given  time  /,  the  unit-stress  should  lie  between  S 
and  25.  Hence  there  must  exist  a  certain  function  <f>(i)  so  that 
the  dynamic  unit-stress  is  given  by  T  =  <f>(t)S;  when  /  is  large, 
T  must  equal  5;  when  /  is  zero,  T  will  be  26*.  Many  empirical 
expressions  can  be  derived  which  satisfy  these  limiting  condi- 
tions, but  the  determination  of  the  theoretic  form  of  <j>(t)  is  greatly 


ART.  136  STRESSES  DUE   TO   LlVE   LOADS  351 

to  be  desired,  because  a  knowledge  of  it  would  be  of  much 
practical  benefit  in  promoting  the  correct  design  of  members  of 
bridge  trusses. 

The  discussion  published  in  Zimmermann,  in  1896,  regarding 
the  increase  in  stress  and  deflection  due  to  the  velocity  of  a  live 
load  when  crossing  a  simple  beam,  is  probably  the  nearest  approach 
to  the  solution  of  this  important  problem,  but  the  formulas  deduced 
are  too  complicated  to  be  given  here.  Let  v  be  the  velocity 
of  a  single  load  P  which  rolls  over  the  beam  of  span  /  and  depth  d, 
and  let  /  be  the  static  deflection  due  to  P.  When  v  is  zero,  the 
dynamic  unit-stress  is  S'}  as  v  increases  the  dynamic  unit-stress 
increases,  but  it  can  never  become  as  great  as  2$ ;  when  v2  has 
the  value  gl2/&f,  where  g  is  the  acceleration  of  gravity,  the  load  P 
reaches  the  middle  of  the  span  in  the  same  time  that  gravity 
causes  a  body  to  fall  freely  through  the  distance  /,  then  also 
T  =  Sm,  when  v  has  a  greater  value,  then  T  is  less  than  5;  when 
v2  =  gr,  where  r  is  the  radius  of  the  earth,  then  T  =  o.  The  impor- 
tant cases  hence  occur  when  v  is  less  than  (gl2/8f)*.  From  Zim- 
merman's investigation,  there  may  be  written, 

T=S(i-  2/?)/(i  -  s/?)  /?  =  8Sv2/3Edg 

which  applies  only  when  /?  is  less  than  o.i,  but  this  covers  most 
cases  of  usual  speeds  of  live  loads  on  beams.  For  example,  take 
a  stringer  in  a  bridge  floor  which  is  2  feet  deep;  the  ratio  E/S 
is  about  4000,  and  hence  /?  =  0.082  for  a  velocity  of  60  miles 
per  hour  or  88  feet  per  second ;  then  T=I.II  so  that  the  dynamic 
stress  is  ii  percent  greater  than  the  static  stress. 

While  the  above  theoretic  formula  gives  much  lower  values 
of  T  than  those  used  in  bridge  practice,  it  may  be  noted  that 
it  refers  only  to  the  middle  of  the  span  of  the  beam,  and 
that  for  other  points  the  theoretic  percentage  of  increase  may 
be  much  greater.  For  the  quarter  points  of  a  short  span  under 
speeds  varying  from  80  to  100  miles  per  hour,  the  investigations 
of  Zimmerman  indicate  that  the  percentage  may  be  as  high 
as  65  percent.  The  empirical  percentages  used  in  bridge  prac- 
tice range  from  100  to  50  percent,  so  that  it  is  plain  that  these 


352  IMPACT  AND  FATIGUE  CHAF.  XIV 

are  not  too  large,  particularly  when  it  is  considered  that  they 
include  the  effect  of  the  shocks  due  to  the  hammer  of  wheels 
which  are  not  properly  balanced. 

Prob.  1360.  Deduce  the  condition  that  the  load  P  shall  reach  the 
middle  of  the  span  in  the  same  time  that  a  body  falls  freely  by  gravity 
through  the  deflection  /. 

Prob.  1366.  Consult  Zimmermann's  Schwingungen  eines  Tragers 
mit  bewegter  Last  (Leipzig,  1896),  and  show  that  the  above  formula 
agrees  with  the  one  given  by  him  on  page  39. 

ART.  137.     FATIGUE  OF  MATERIALS 

The  ultimate  strength  Su  is  usually  understood  to  be  that 
steady  unit-stress  which  causes  rupture  of  a  bar  at  one  appli- 
cation. Experience  and  experiments,  however,  teach  that  rup- 
ture may  be  caused  by  a  unit-stress  somewhat  less  than  Su  when 
it  is  applied  a  sufficient  number  of  times  to  a  bar.  The  experi- 
ments made  by  Wohler  from  1859  to  1870  were  the  first  that 
indicated  the  laws  which  govern  the  rupture  of  metals  under 
repeated  applications  of  stress.  For  instance,  he  found  that 
the  rupture  of  a  bar  of  wrought  iron  by  tension  was  caused  in 
the  following  different  ways : 

by  800  applications  of  52  800  pounds  per  square  inch 

by  107  ooo  applications  of  48  400  pounds  per  square  inch 

by  450  ooo  applications  of  39  ooo  pounds  per  square  inch 

by  10  140  ooo  applications  of  35  ooo  pounds  per  square  inch 

The  range  of  stress  in  each  of  these  applications  was  from  o  to 
the  designated  number  of  pounds  per  square  inch.  Here  it  is 
seen  that  the  breaking  unit-stress  decreases  as  the  number  of  ap- 
plications increase.  In  other  experiments  where  the  initial  stress 
was  not  o,  but  a  permanent  value  S,  the  same  law  was  seen  to 
hold  good.  It  was  further  observed  that  a  bar  could  be  strained 
from  o  up  to  a  stress  near  its  elastic  limit  an  enormous  number 
of  times  without  rupture,  and  it  was  also  found  that  a  bar  could 
be  ruptured  by  a  stress  less  than  its  elastic  limit  under  a  large 
number  of  repetitions  of  stress  which  alternated  from  tension 
to  compression  and  back  again. 


ART.  137  FATIGUE    OF   MATERIALS  353 

Wohler's  experiments  were  made  on  repeated  tensile  stresses, 
repeated  flexural  stresses,  and  on  flexural  stresses  alternating 
from  tension  to  compression,  these  being  produced  by  a  machine 
which  brought  repeated  loads  upon  the  specimen  for  long  periods 
of  time,  as  high  as  forty  millions  of  repetitions  being  made  in 
some  cases.  Similar  experiments  were  later  made  by  Bauschinger 
and  others  on  steel,  and  from  the  recorded  results  the  following 
laws  may  be  stated: 

1.  The  rupture  of  a  bar  may  be  caused  by  repeated  appli- 
cations of  a  unit-stress  less  than  the  ultimate  strength  of  the 
material. 

2.  The  greater  the  range  of  stress,  the  less  is  the  unit-stress 
required  to  produce  rupture  after  an  enormous  number  of 
applications. 

3.  When  the  unit-stress  in  a  bar  varies  from  o  up  to  the 
elastic  limit,  an  enormous  number  of  applications  is  required 
to  cause  rupture. 

4.  A  range  of  stress  from  tension  into  compression  and 
back  again,  produces  rupture  with  a  less  number  of  applica- 
tions than  the  same  range  in  stress  of  one  kind  only. 

5.  When  the  range  of  stress  in  tension  is  equal  to  that  in 
compression,  the  unit-stress  that  produces  rupture  after  an 
enormous  number  of  applications  is  a  little  greater  than  one- 
half  the  elastic  limit. 

The  term  '  enormous  number '  means  about  40  millions, 
that  being  roughly  the  number  used  by  Wohler  to  cause  rupture 
under  the  conditions  stated.  For  all  cases  of  repeated  stress  in 
bridges,  this  great  number  will  not  be  exceeded  during  the  natu- 
ral life  of  the  structure;  for  locomotive  axles  and  moving  parts 
of  machines,  however,  a  larger  number  of  repetitions  of  stress 
may  occur. 

The  word  '  fatigue '  means  the  loss  of  molecular  strength 
tinder  stresses  often  repeated.  When  a  bar  is  stressed  above  the 
elastic  limit  its  temperature  increases  due  to  internal  molecular 
friction  (Art.  147)  and  it  is  known  that  the  elastic  properties 
of  the  material  are  injured.  Hence  in  a  general  way  it  is  easy 
to  explain  why  fatigue  occurs  under  repeated  stressses  that  exceed 


354  IMPACT  AND  FATIGUE  CHAP,  xiv 

the  elastic  limit.  An  examination  of  fractures  of  bars  after  an 
enormous  number  of  repetitions  shows  certain  small  surfaces 
where  sliding  or  shearing  has  occurred;  these  are  called  'micro- 
flaws,'  although  they  can  often  be  seen  without  the  use  of  a  micro- 
scope. In  iron  and  steel  these  flaws  begin  along  the  surfaces  of 
the  ferrite  crystals  and  later  are  extended  to  cause  cracks  along 
the  cleavage  planes  of  the  crystals. 

When  the  elastic  limit  is  not  exceeded  it  is  not  so  easy  to  under- 
stand why  fatigue  should  occur  under  repeated  stresses.  How- 
ever, physical  and  thermodynamic  discussions  have  proved  that 
small  changes  in  temperature  occur  when  a  bar  of  metal  is  stressed 
within  the  elastic  limit,  it  becoming  cooler  under  tension  and 
warmer  under  compression.  The  measurements  of  these  changes 
made  by  Turner,  in  1902,  have  shown  that  these  changes  in  tem- 
peratures continue  at  a  uniform  rate  up  to  about  three-fifths 
of  the  elastic  limit,  and  that  then  a  marked  change  occurs,  the 
bar  under  tension  then  beginning  to  grow  warmer  while  the 
temperature  of  the  bar  under  compression  increases  at  a  more 
rapid  rate.  It  thus  appears  that  for  stresses  higher  than  about 
three-fifths  of  the  elastic  limit,  at  least,  energy  is  converted  into 
heat  under  repeated  applications;  probably  this  occurs  also  at 
lower  stresses  when  repeated  stresses  range  from  tension  into 
compression  in  a  bar,  or  when  a  beam  is  subject  to  alternating 
flexure.  The  valuable  experiments  of  Turner  hence  throw  light 
upon  the  reason  why  fatigue  occurs  under  alternating  stresses, 
and  it  is  likely  that  further  investigations  in  this  direction  may 
lead  to  other  important  conclusions.  The  discussions  in  Arts. 
146-147  indicate  that  internal  friction  occurs  under  stresses  that 
do  not  exceed  the  elastic  limit,  and  this  point  of  view  is  also  one 
which  will  assist  future  investigations. 

In  Art.  7  it  was  recognized  that  allowable  unit-stresses  should 
be  less  for  bars  subject  to  varying  loads  than  for  those  carrying 
steady  loads  only.  It  has  indeed  long  been  the  practice  of  de- 
signers to  grade  the  allowable  working  stresses  for  bars  according 
to  the  range  of  stress  to  which  they  might  be  liable  to  be  sub- 
jected. The  above  laws  of  fatigue  furnish  a  method  of  doing 


ART.  138  STRENGTH    UNDER   FATIGUE  355 

this  which  has  been  used  by  some  engineers,  and  formulas  for 
that  purpose  will  be  deduced  in  the  next  article. 

Prob.  137a.  How  many  years  will  probably  be  required  for  a  loco- 
motive axle  to  receive  forty  million  repetitions  of  flexural  stresses  ? 

Prob.  1376.  Consult  Turner's  paper  in  Transactions  American 
Society  of  Civil  Engineers,  1902,  Vol.  48,  and  examine  the  thermal 
stress  curves  derived  from  his  experiments. 

ART.  138.    STRENGTH  UNDER  FATIGUE. 

Consider  a  bar  in  which  the  unit-stress  varies  from  S'  to  S, 
the  latter  being  the  greater  numerically.  Both  S'  and  S  may 
be  tension  or  both  may  be  compression,  or  one  may  be  tension 
and  the  other  compression;  in  the  last  case  the  sign  of  S'  is  to 
be  taken  as  minus.  Consider  the  stress  to  be  repeated  an  enor- 
mous number  of  times  from  S'  to  5,  and  rupture  to  then  occur 
under  the  greater  unit-stress  S,  which  may  be  called  the  strength 
of  the  material  under  fatigue.  By  the  second  law  above  stated 
S  is  some  function  of  S  —  S' ;  this  is  equivalent  to  saying  that 
S  is  a  function  of  S(i—  S'/S),  or  more  simply  a  function  of  S'/S. 
Now  if  P'  and  P  be  the  total  stresses  on  the  bar,  the  ratio  S'/S 
equals  P'/P,  and  hence  the  unit-stress  S  which  causes  rupture 
after  an  enormous  number  of  repetitions  is  a  function  of  P' 'IP. 

A  formula  for  5  when  the  limiting  stresses  Pr  and  P  are  both 
tension  or  both  compression,  so  that  P'/P  is  always  positive, 
was  deduced  by  Launhardt  in  1873.  Let  the  values  of  this  ratio 
be  taken  as  abscissas  ranging  from  o  to  i,  while  those  of  S  are 
ordinates,  as  in  Fig.  1380.  Let  the  function  of  P'/P  be  supposed 
to  represent  a  straight  line  which  has  the  equation  S  =  Ci  +  Cz(P' '/P) 
in  which  Ci  and  C2  are  constants  to  be  determined.  Let  Su 
be  the  ultimate  strength  of  the  material  and  Se  the  elastic  limit. 
Now  if  P'/P  is  unity,  then  5  is  Su  and  hence  Ci  +C2  =SU\  also, 
from  the  third  law  of  the  last  article,  S  is  Se  when  P'/P  is  zero. 
These  two  conditions  give  C\  =  Se  and  C2=Su—Se,  and  the  equa- 
tion of  the  straight  line  becomes, 

S=Se+(S*-S.)P'/P  or  s= 


356  IMPACT  AND  FATIGUE  CHAP.  XIV 

which  is  Launhardt's  formula  for  the  unit-stress  5  that  ruptures 
the  bar  after  an  enormous  number  of  repetitions  of  a  load  that 
ranges  from  P'  to  P.  For  structural  steel,  using  the  mean  values 
of  Su  and  Se  in  Tables  2  and  3,  this  becomes  S  =  35  000(1  +  -?P'/P) ; 
for  wrought  iron  it  becomes  S  =  25  ooo(i+P'/P).  For  example, 
let  a  bar  of  structural  steel  range  in  tension  from  80  ooo  to  160  ooo 
pounds;  then  P'/P  is  0.5,  and  5=47  5°°  pounds  per  square 
inch  is  the  unit-stress  that  will  cause  rupture  after  an  enormous 
number  of  repetitions. 

A  formula  for  S  when  the  bar  ranges  in  stress  from  Pf  to  P, 
one  being  tension  and  the  other  compression,  and  P  being  the 
greater  numerically,  was  deduced  by  Weyrauch  in  1877.  Here 
P'/P  is  always  negative,  and  the  law  connecting  it  with  S  is 
again  assumed  to  be  5  =  Ci  +  C2(P'/P);  Fig.  1386  represents  this 
case.  Let  5^  be  the  unit-stress  at  the  elastic  limit  and  Sa  the 
unit-stress  which,  under  the  fifth  law  of  the  last  article,  causes 
rupture  when  the  load  alternates  from  a  certain  value  in  tension 
to  the  same  value  in  compression.  By  the  third  law,  if  P'/P  is 
zero,  then  S  is  Se  and  hence  Ci  =  5,.  By  the  fifth  law,  if  P'/P  is 
—  i,  then  S  is  Sa  and  hence  C2=Se  —  Sa.  The  equation  of  the 
straight  line  therefore  becomes, 

S  =  S,+  (Se-Sa)P'/P  or  S= 


which  is  Weyrauch's  formula  for  the  unit-stress  which  ruptures 
a  bar  after  an  enormous  number  of  repetitions  of  a  load  alter- 
nating from  tension  to  compression  and  back  again.  Sa  is 
usually  taken  as  $Se  in  the  absence  of  knowledge  regarding  its 
exact  value.  For  structural  steel  the  formula  becomes  5  = 
35  ooo  (i  +  JP'/P),  in  which  P'/P  is  always  negative.  Thus,  if 
Pf  is  80  ooo  pounds  compression  and  P  is  160  ooo  pounds  ten- 
sion, then  P'/P=  —  0.5,  and  5  =  26200  pounds  per  square  inch 
is  the  unit-stress  that  will  cause  rupture. 

Another  formula,  deduced  by  the  author  in  1884,  gives  values 
of  5  for  both  positive  and  negative  values  of  P'/P,  and  thus 
includes  the  two  cases  discussed  above.  The  law  of  variation 
of  5  is  assumed  to  be  represented  by  a  curve  joining  the  tops  of 


ART.  138 


STRENGTH  UNDER  FATIGUE 


357 


the  three  ordinates  Su,  Se,  Sa,  in  Fig.  l3Sc.  The  simplest  curve 
is  a  parabola  given  by  the  equation  S  =  Ci  +C2(P'/P)  +  C3(P'/P)2. 
To  determine  the  three  constants,  consider,  first  that  5  becomes 
Su  when  P'/P=-fi,  and  hence  C\  +  C2  +  C3  =  SU;  secondly, 
that  5  becomes  Se  when  P'/P=o,  and  hence  Ci=Se\  thirdly, 
that  S  becomes  Sa  when  P'/P  =  -i,  and  hence  d  -C2  +  C3  =Sa. 
From  these  conditions,  the  values  of  Ci,  C2,  C3  are  found  and 
the  equation  of  the  parabola  becomes, 

P'  /P'X2 

which  is  a  formula  for  the  rupturing  unit-stress  S  when  the  total 
stress  ranges  an  enormous  number  of  times  from  P'  to  P.  When 
P'  and  P  are  both  tension  or  both  compression,  the  ratio  P'/P 
is  positive;  when  one  is  tension  and  the  other  compression,  P'/P 
is  negative.  It  is  seen  that  (138)  gives  values  of  5  a  little  smaller 
rhan  those  found  from  the  straight-line  formulas. 


For  structural  steel,  where  Su  =  60000,  5*  =  35000,  and 
Sa  =  ifj  500  pounds  per  square  inch,  the  formula  (138)  reduces  to, 

5  =  35  ooo[i +0.61  (P'/P) +  0.1 1  (P'/P)2] 

For  a  bar  of  such  steel  in  which  the  stress  ranges  from  180  ooo 
pounds  tension  under  dead  load  to  540  ooo  pounds  tension  under 
live  load,  the  value  of  P'/P  is  +J,  and  the  formula  gives  5  = 
42  500  pounds  per  square  inch.  If  the  stress  ranges  from 
180000  pounds  compression  to  540000  pounds  tension,  then 
P'/P  is  —  J,  and  5  =  28  300  pounds  per  square  inch. 

When  the  above  formulas  are  used  for  designing,  a  factor 
of  safety  is  applied,  the  computed  values  of  S  being  divided 
by  this  factor,  and  thus  the  allowable  unit-stress  is  obtained. 
About  1880  the  formula.?  of  Launhardt  and  Weyrauch  were 


358  IMPACT  AND  FATIGUE  CHAP,  xiv 

extensively  used  in  determining  the  allowable  unit-stresses  for 
designing  members  of  bridge  trusses,  but  their  use  has  gradu- 
ally been  replaced  in  the  United  States  of  America  by  other 
methods.  Since  no  unit-stress  used  for  a  bridge  member  can 
be  allowed  to  be  greater  than  about  one -half  of  the  elastic  limit 
of  the  material,  it  is  claimed  by  many  engineers  that  the  ideas 
of  fatigue  cannot  enter  in  making  the  design.  Nevertheless 
this  question  must  not  be  ignored,  especially  for  locomotive 
axles  and  tires  and  for  parts  of  machines  subject  to  shocks.  Tests 
of  materials  under  repeated  stresses,  or  endurance  tests  as  they 
are  sometimes  called,  are  still  in  progress  at  the  Watertown 
Arsenal,  at  the  testing  laboratory  of  the  Pennsylvania  Railroad, 
and  in  other  places;  when  sufficient  records  have  been  accumu- 
lated they  will  prove  of  great  value  in  further  investigations  into 
this  subject.- 

Prob.  138a.  A  short  bar  of  wrought  iron  is  subject  to  repeated 
stresses  ranging  from  16  ooo  pounds  compression  to  80  ooo  pounds 
tension.  What  should  be  the  area  of  its  cross-section  for  a  factor 
of  safety  of  5  ? 

Prob.  1386.  Consult  Tests  of  Metals,  published  annually  by  the 
ordnance  office  of  the  U.  S.  Army,  and  describe  some  of  the  endurance 
tests  on  rotating  shafts  made  by  Howard. 


ART.  139  PRINCIPLES  AND    LAWS  359 

CHAPTER  XV 
TRUE  INTERNAL  STRESSES 

ART.  139.    PRINCIPLES  AND  LAWS 

In  Art.  13  it  was  explained  that  a  bar  under  tension  suffers 
a  contraction  in  its  section  area,  each  lateral  dimension  having 
a  unit-contraction  proportional  to  the  longitudinal  unit-elonga- 
tions, when  the  elastic  limit  of  the  material  is  not  exceeded. 
Let  S  be  the  tensile  unit-stress,  e  the  unit-elongation,  X  the  fac- 
tor of  lateral  contraction,  and  E  the  modulus  of  elasticity  of  the 
material;  the  lateral  unit-shortening  is  then  As.  Since  S  =  eE 
is  the  relation  between  S  and  e  (Art.  9),  it  may  be  considered 
that  the  lateral  unit-shortening  As  corresponds  to  a  unit-stress  T 
which  has  such  a  value  that  T  =  AeE,  where  T  is  a  compressive 
unit-stress  which  would  produce  the  unit-shortening  As  in  the 
absence  of  any  axial  stress.  Thus,  T  =  AS  is  called  a  true  internal 
stress  which  acts  as  a  compression  at  right  angles  to  the  axis 
of  the  bar. 

The  mean  value  of  A  for  wrought  iron  and  steel  is  about  J. 
Accordingly,  a  steel  bar  under  the  tensile  unit-stress  S  suffers 
a  true  internal  compressive  unit-stress  of  JS  in  all  directions  at 
right  angles  to  its  length;  similarly,  a  steel  bar  under  the  com- 
pressive unit-stress  5  suffers  a  true  internal  tensile  unit-stress 
of  JS  in  all  directions  at  right  angles  to  its  length.  For  instance, 
let  a  steel  bar  2X3  inches  in  section  and  10  inches  long  be  sub- 
ject to  a  tension  of  90  ooo  pounds ;  the  axial  tensile  unit-stress  S 
is  15  ooo  pounds  per  square  inch  and  the  lateral  internal  com- 
pressive unit-stress  is  5  ooo  pounds  per  square  inch.  The  same 
lateral  deformation  of  the  bar,  when  no  axial  load  is  acting,  might 
be  produced  by  two  compressive  loads  acting  at  right  angles  to 
each  other,  one  uniformly  distributed  over  the  side  of  20  square 
inches  area  and  the  other  over  the  side  of  30  square  inches  area ; 
it  may  be  shown  from  the  following  discussion  that  these  two 
compressive  loads  are  150  ooo  and  225  ooo  pounds. 


360  TRUE  INTERNAL  STRESSES  CHAP.  XV 

When  applied  tensile  forces  act  upon  a  body  in  three  directions, 
each  force  being  at  right  angles  to  the  plane  of  the  other  two, 
there  is  an  elongation  due  to  each  force  in  its  own  direction  and 
a  shortening  in  directions  normal  to  it.  It  is  a  reasonable  assump- 
tion that  each  force  produces  its  deformations  independently  of 
the  other  two,  and  this  is  also  justified  by  experience  and  experi- 
ment. The  true  stress  in  any  direction  depends  upon  the  actual 
deformation  in  its  direction.  The  letter  S  will  denote  the  apparent 
unit-stress  as  computed  by  the  methods  of  the  preceding  chapters, 
while  T  will  denote  the  true  unit-stress  corresponding  to  the 
actual  deformation.  The  injury  done  to  a  body  does  not  depend 
upon  the  actual  stress  or  pressure  but  upon  the  actual  deforma- 
tions produced,  and  the  true  stresses  are  those  corresponding 
to  these  deformations. 

Let  a  homogeneous  parallelepiped  be  subject  to  tensile  forces 
acting  normally  upon  its  six  faces,  those  upon  opposite  faces  being 

equal.  Let  the  edges  of  the  parallel- 
epiped be  designated  by  01,  02,  03, 
as  in  Fig.  139.  Let  Si  be  the  normal 
unit-stress  upon  the  two  faces  per- 
pendicular to  the  edge  01,  and  £2  and 
S3  those  upon  the  faces  normal  to  02 
and  03;  thus  the  directions  of  Si,  Sa> 
S3  are  parallel  to  01,  02,  03,  respectively.  Then,  supposing  that 
the  modulus  of  elasticity  E  and  the  factor  of  lateral  contraction  / 
are  the  same  in  all  directions,  the  true  unit -elongations  £1,  £2,  £3 
in  the  three  directions  are  found  from  the  expressions, 


Now  E£I maybe  designated  by  T\,  this  being  the  unit-stress  which 
would  produce  the  elongation  si  in  the  direction  01  if  82  and  S3 
were  not  acting;  also  the  unit-stresses  Ee2  and  Ee3  may  be  desig- 
nated by  7*2  and  TV  Hence  it  follows  that 

Tl=S1-*S2-*S3      T2  =  S2-te3-XSl       T3  =  S3-lSl-lS2     (139) 

are  the  true  stresses  acting  in  the  three  rectangular  directions. 
If  any  stress  S  is  compression,  it  is  to  be  taken  as  negative  in  the 


A.RT.  139  PRINCIPLES  AND  LAWS  361 

formulas,  and  then  the  true  stresses  are  tensile  or  compressive 
according  as  their  numerical  values  are  positive  or  negative. 

For  example,  let  a  cube  be  stressed  upon  all  sides  by  the 
apparent  unit-stresses  S;  then  the  true  internal  unit-stress  T 
is  5(i— 2A);  for  steel  A  =  J,  and  T  =  ^S,  and  thus  the  linear  defor- 
mation is  only  one-third  of  that  due  to  a  unit-stress  6*  applied 
upon  two  opposite  faces.  Again,  if  a  bar  has  a  tension  Si  in 
the  direction  of  its  length,  and  no  stresses  upon  its  sides,  then 
Ti  =5i,  while  T2  =  T3=  -ASi- 

As  a  simple  example,  let  a  steel  bar  2  feet  long  and  3X2 
inches  in  section  area  be  subject  to  a  tension  of  60  ooo  pounds 
in  the  direction  of  its  length  and  to  a  compression  of  432  ooo 
pounds  upon  the  two  opposite  flat  sides.  Here  Si  =  60  000/6 
=  10  ooo  pounds  per  square  inch,  $2  =  —432  000/72  =  —  6  ooo 
pounds  per  square  inch,  and  S3  =o.  Then  from  (139),  taking 
A  as  J,  the  true  internal  stresses  are  TI  =  + 1 2  ooo,  T%  =  —  9  330, 
TZ  =  —  i  330  pounds  per  square  inch,  and  it  is  thus  seen  that  the 
true  tensile  unit-stress  is  20  percent  greater  than  the  apparent, 
while  the  true  compressive  unit-stress  is  more  than  50  percent 
greater  than  the  apparent. 

The  term  '  apparent  stresses '  will  be  used  to  indicate  the 
stresses  computed  by  the  methods  of  the  previous  chapters  where 
no  lateral  deformation  has  been  taken  into  account.  In  Chapter 
XI  such  stresses  have  been  combined  in  order  to  obtain  the 
resultant  maximum  tension,  compression,  and  shear,  but  it  will 
now  be  shown  that  the  true  internal  stresses  corresponding  to 
the  actual  deformations  of  the  material  are  often  much  greater 
than  the  apparent  ones.  It  is  very  important  to  consider  these 
true  stresses  in  many  problems  of  investigation  and  design  which 
occur  in  engineering  practice. 

Prob.  139.  A  common  brick,  2^X4X8^  inches  in  size,  is  subject 
to  a  compression  of  3  200  pounds  upon  its  top  and  bottom  faces,  500 
pounds  upon  its  sides,  and  60  pounds  upon  its  ends.  Taking  X  as 
0.2,  compute  the  true  internal  stresses  in  the  three  directions. 


362  TRUE  INTERNAL  STRESSES  CHAP.  XV 

ART.  140.     SHEAR  DUE  TO  NORMAL  STRESS 

The  term  *  normal  stress '  is  used  for  the  tension  or  compression 
that  acts  normally  to  a  plane  in  the  interior  of  a  body.  The 
rectangular  bar  in  Fig.  1400  may  be  said  to  be  acted  upon  by 
normal  loads,  and  planes  perpendicular  to  these  loads  are  said  to 
be  subject  to  normal  unit-stress.  Other  normal  stresses  also  act 
upon  other  planes  within  the  bar,  but  it  will  be  shown  that  the 
normal  unit-stresses  upon  such  planes  are  less  than  upon  planes 
perpendicular  to  the  directions  of  PI  and  P2.  Let  Si  be  the  normal 
unit-stress  on  a  plane  perpendicular  to  PI,  and  52  that  on  a  plane 
perpendicular  to  P2;  then  the  true  unit-stresses  TI  and  T2,  as 
also  the  true  unit-stress  T%  at  right  angles  to  these,  are  readily 
found  by  the  methods  of  Art.  139.  There  also  exist  shearing 
stresses  in  the  bar  which  will  now  be  considered.  Let  any  plane 
be  drawn  cutting  it  obliquely  and  let  the  given  forces  be  resolved 
into  components  parallel  to  this  plane;  the  sum  of  these  com- 
ponents forms  a  shear  acting  along  the  plane,  and  the  intensity 
of  the  shear  will  vary  with  the  inclination  of  the  plane.  It  is 
required  to  find  the  maximum  shearing  unit-stresses. 

Let  /  be  the  length,  b  be  the  breadth,  and  d  the  depth  of  the 
rectangular  bar  in  Fig.  140a,  subject  to  the  two  normal  forces 
PI  and  P2,  while  there  is  no  force  acting  upon  the  side  whose 
area  is  Id.  The  normal  unit-stresses  then  are  Si=Pi/bd,  S2  = 
P2/bl,  S3=o.  Let  Fig.  1406  represent  any  elementary  parallelo- 
piped  in  the  interior  of  the  bar,  having  the  length  dxt  depth  dy, 
and  width  unity;  then  Sidy  is  the  normal  stress  upon  its  ends, 
and  S2dx  is  the  normal  stress  upon  the  upper  and  lower  sides. 
Let  0  be  the  angle  which  the  diagonal  dz  makes  with  dx,  and 
let  S'  be  the  shearing  unit-stress  that  acts  along  the  diagonal. 
The  total  shearing  stress  along  the  diagonal  then  is  S'dz  and  this 
is  equal  to  the  algebraic  sum  of  the  components  of  the  normal 
stresses  in  that  direction.  Accordingly,  noting  that  dy/dz  = 
sin#  and  dx/dz  =  cosO,  there  result, 

S'dz=Sidy  .  cos6-S2doc  sin/?          or          S'  =  (Si-S2)sm  6  cos  6 
and   the   second   equation    gives  the  shearing  unit-stress   alone 


ART.  140 


SHEAR  DUE  TO  NORMAL  STRESS 


363 


any  plane  which  makes  the  angle  6  with  the  direction  of  S\. 
The  maximum  value  of  S'  occurs  for  6=  +45°  or  6=  -45°,  and 
for  both  cases  may  be  written  S'  =  ±  J(Si  —  82),  that  is,  the  maxi- 
mum shearing  unit-stress  occurs  on  two  planes  which  bisect  the 
directions  of  Si  and  £2,  and  its  numerical  value  is  equal  to  one- 
half  of  their  difference.  The  broken  lines  in  Fig.  1400  show 
these  two  sets  of  planes. 

ts25x 


v 

,\ 

V     X' 

x 

> 

<  X 

'  >(  X 

Sx\ 

X 

X  ^ 

V 

Sidy 


Fig.  1400 


Fig.  1406 


The  same  conclusion  follows  when  true  stresses  are  con- 
sidered. If  TI  and  T2  are  the  true  unit-stresses  due  to  the 
apparent  unit-stresses  Si  and  5*2,  then  the  true  maximum  shear- 
ing unit-stresses  are  equal  to  one-half  their  difference,  and  they 
act  in  planes  which  bisect  the  directions  of  TI  and  T2.  Accord- 
ingly, the  formulas, 

5'  =  J(5i-52)  and  T'  =  %(Ti-T2}  (140) 

give  the  maximum  internal  shearing  unit-stresses.  These  may 
be  either  positive  or  negative,  but  it  is  best  to  consider  internal 
shear  as  a  signless  quantity,  since  it  acts  in  opposite  directions 
on  opposite  sides  of  the  plane. 

As  a  numerical  example,  take  a  cast-iron  bar  for  which  the 
factor  of  lateral  contraction  A  is  J;  let  it  be  one  square  inch  in 
section  area  and  be  subject  only  to  an  axial  tension  of  2  400 
pounds.  Then  Si  =  +2  400  and  52=o,  whence  the  maximum 
apparent  shearing  unit-stress  is  S'  =  i  200  pounds  per  square 
inch.  From  (139)  the  true  axial  unit-stress  is  T\  =  +  2  400  and 
the  true  lateral  unit-stress  is  T2=  -600  pounds  per  square  inch. 
Accordingly  the  maximum  true  shearing  unit-stress  is  T'  =  i  500 
pounds  per  square  inch,  which  is  25  percent  greater  than  the 
apparent.  It  is  indeed  very  common  to  find  that  the  true  stresses 
based  on  the  actual  deformations  are  much  larger  than  the  stresses 
computed  from  the  common  theory,  and  this  is  one  reason  for 
the  use  of  high  factors  of  safety. 


364  TRUE  INTERNAL  STRESSES  CHAP.  XV 

The  above  discussion  applies  equally  well  when  one  or  both 
of  the  applied  loads  is  compression.  For  example,  let  the  axial 
unit-stress  Si  be  tension  and  the  lateral  unit-stress  £2  be  com- 
pression, each  equal  to  2  400  pounds  per  square  inch.  Then 
the  apparent  maximum  shearing  unit-stress  is  S'  =  J(2  400  +  2  400) 
=  -2  400  pounds  per  square  inch.  For  \  =  J,  the  true  axial  stress 
is  TI  =  +3  200  and  the  true  lateral  stress  is  T2  =  —  3  200,  so  that 
the  true  maximum  shearing  unit-stress  is  7"'=  3  200  pounds  pel 
square  inch,  which  is  33  percent  higher  than  S'. 

When  Si  and  £2  are  equal  numerically,  both  being  tension 
or  both  compression,  then  S' =o,  and  also  T"'=o;  that  is,  a 
parallelepiped  under  uniform  stress  in  two  rectangular  directions 
has  no  internal  shearing  stress.  The  same  is  true  when  a  body 
is  acted  upon  by  equal  tensions  or  pressures  in  three  rectangular 
directions,  for  the  third  stress  Ss  exerts  an  equal  influence  upon 
the  two  normal  to  it. 

When  there  are  three  unit-stresses  Si,  S2,  Ss,  acting  upon 
a  paralellopiped  in  three  rectangular  directions,  the  shearing 
unit-stress  on  a  plane  parallel  to  Si  and  S2  is  not  influenced  by 
Ss,  and  hence  J(Si—  S2)  is  the  maximum  shearing  unit-stress 
for  such  a  plane.  Similarly,  J(Si  —  Ss)  and  £(£2  —  Ss)  are  the 
maximum  shearing  unit-stresses  for  planes  parallel  to  Si  and  S2 
and  to  S2  and  Ss  respectively.  The  same  holds  true  for  the 
true  stresses  TI,  T%,  T$\  an  algebraic  discussion  of  this  case 
will  be  found  in  Art.  178.  As  an  example,  let  a  rectangular 
bar  be  subject  to  an  axial  tension  of  3  ooo  pounds  per  square 
inch,  and  to  a  compression  of  6  ooo  pounds  per  square  inch  upon 
two  opposite  sides.  Here  Si  =  +3  ooo,  S2  =  —  6000,  Ss=o, 
and  hence  the  three  maximum  apparent  shearing  stresses  are 
4  500,  3  ooo,  i  500  pounds  per  square  inch.  But  from  (139), 
taking  A  as  J,  the  true  axial  and  lateral  stresses  are  T\  =  +  5  coo, 
7*2  =  —  7  ooo,  T3  =  + 1  ooo,  whence  the  three  maximum  true 
shearing  stresses  are  6  ooo,  4  ooo,  2  ooo  pounds  per  square  inch. 
Here  the  true  axial  stress  is  67  percent  greater  than  the  apparent, 
while  the  true  shearing  stresses  are  33  percent  greater  than  the 
apparent  ones. 


ART.  141         COMBINED  SHEAR  AND  AXIAL  STRESS  365 

Prob.  140.  Compute  the  maximum  shearing  unit-stresses,  both 
apparent  and  true,  for  the  data  given  in  Problem  139. 

ART.  141.     COMBINED  SHEAR  AND  AXIAL  STRESS 

Formulas  were  deduced  in  Art.  105  for  the  maximum  apparent 
stresses  of  tension,  compression,  and  shear,  due  to  the  simul- 
taneous action  of  an  axial  load  and  a  cross-shear.  It  was  shown 
that  there  are  two  planes  at  right  angles  to  each  other  upon  which 
there  are  no  shearing  stresses,  one  being  under  normal  tension  Si 
and  the  other  under  normal  compression  S2.  Let  S  be  a  given 
axial  unit-stress  of  tension  and  Ss  the  shearing  unit-stress  acting 
at  right  angles  to  it.  Then  the  formulas  give  the  following  values 
of  the  maximum  tensile  stress  Si,  the  maximum  compressive  unit- 
stress  S2,  and  the  maximum  shearing  unit-stress  S', 

S1==^+(Ss2+iS2)*      S2  =  iS-(Ss2+i?2)*      S'=(SS2+$S2)*    (141) 

It  is  here  seen  that  the  value  of  S'  is  the  same  as  that  of  ^(Si~  S2). 
Hence  when  Si  and  £2  have  been  computed,  ths  subsequent  dis- 
cussion is  exactly  like  that  of  the  last  article.  When  S  is  tension, 
as  above  considered,  Si  is  tension  and  S2  is  compression  ;  when  S 
is  compression,  Si  is  compression  and  £2  is  tension. 

Let  X  be  the  factor  of  lateral  contraction,  and  T\  and  T2  the 
true  internal  unit-stresses  corresponding  to  Si  and  -5*2.  Then, 
by  (139),  the  value  of  Tl  is  Si-XS2  and  that  of  T2  is  S2-*Si; 
substituting  in  these  the  above  values  of  Si  and  S2)  there  are  found, 

I 
j 

from  which  7\  and  T2  may  be  directly  computed.  For  steel  the 
mean  value  of  A  is  J,  and  hence  for  this  material, 


are  the  true  maximum  tensile  and  compressive  unit-stresses  due  to 
an  axial  unit-stress  S  and  a  shearing  unit-stress  S8  acting  at  right 
angles  to  it.  The  true  maximum  shearing  unit-stress  acts  along 
a  plane  that  bisects  the  directions  of  Si  and  S2  and  its  value  is 
T'=%(Ti-T2).  The  directions  of  TI  and  T2  are  the  same  as 
those  of  Si  and  S2,  and  may  be  found  from  the  expression  for  cot2<£ 


366  TRUE  INTERNAL  STRESSES  CHAP.  XV 

deduced  in  Art.  105,  namely,  cot2<£  =  —  %S/S8,  where  the  two 
values  of  (f>  give  the  angles  included  between  the  direction  of  5 
and  those  of  the  planes  against  which  Si  and  S2  act. 

As  a  numerical  illustration,  take  the  case  of  a  steel  bolt 
subject  to  an  axial  tension  of  2  ooo  and  to  a  cross-shear  of  3  ooo 
pounds  per  square  inch.  Here  S  =  +  2  ooo,  S8  =  3  ooo,  from 
which  Si  =  +4  1 60  and  S2  =  —  2  160  pounds  per  square  inch  are 
the  apparent  maximum  unit-stresses  of  tension  and  compression, 
and  their  directions  are  given  by  cot2^  =  — J.  The  two  values 
of  <f>  then  are  54°  13'  and  144°  .13',  which  show  that  S\  makes  an 
angle  of  35°  47'  and  S2  an  angle  of  54°  13'  with  the  axis  of 
the  bolt.  The  true  unit-stresses  have  the  same  directions  and 
their  values  are  jTi  =  +488o,  T2=  —  3550  pounds  per  square 
inch.  For  the  shearing  unit-stresses  the  maximum  values  are 
S'  =3  160  and  Tr  =4  220  pounds  per  square  inch.  Here  the  true 
maximum  tension  is  17  percent  greater  than  the  apparent,  the 
true  compression  is  64  percent  greater,  and  the  true  shear  is  33 
percent  greater.  There  is  also  a  third  true  compression  T^  =  —  670, 
and  two  other  true  shears  smaller  than  S'  which  act  along  planes 
parallel  to  Si.  It  thus  appears  that  the  actual  internal  stresses 
corresponding  to  the  deformations  of  the  material  are  far  more 
complex  than  and  quite  different  in  value  from  those  computed 
by  the  common  theory. 

The  above  discussion  considers  a  bar  subject  only  to  a  single 
axial  stress  S  and  to  a  cross-shear  Ss.  This  is  a  very  common 
case  in  engineering  practice,  but  other  cases  far  more  complex 
occasionally  occur  where  the  bar  is  subject  to  both  axial  and 
lateral  stresses  and  to  shears  in  different  directions.  The  methods 
of  treating  these  cases  will  be  explained  in  Arts.  177  and  178. 

Prob.  1410.  A  horizontal  bar  of  cast  iron,  2X2X6  inches,  is  under 
an  axial  compression  of  20  ooo  pounds,  and  under  shear  from  a  uniform 
vertical  load  of  8  ooo  pounds  which  rests  upon  it.  Compute  the 
maximum  unit-stresses,  both  apparent  and  true,  and  find  the  direc- 
tions which  they  make  with  the  axis  of  the  bar. 

Prob.  1416.  What  must  be  the  value  of  S8  in  (141)  in  order  that  Si 
and  S2  may  be  equal  ? 


ART.  142  TRUE    STRESSES   IN    BEAMS  367 


ART.  142.    TRUE  STRESSES  IN  BEAMS 

The  first  set  of  formulas  in  the  last  article  furnishes  the  means 
of  ascertaining  the  maximum  apparent  stresses  at  any  point  in 
the  beam,  S  being  the  horizontal  unit-stress  for  that  point  as 
computed  from  the  flexure  formula  and  5S  the  shearing  unit-stress 
as  determined  by  Art.  108.  From  these  the  apparent  unit-stresses 
Si  and  S2  which  act  at  the  given  point  are  computed  and  then 
the  true  unit-stresses  T\  and  T2.  The  discussion  in  the  last  article 
also  shows  that  the  directions  of  T\  and  T2  are  the  same  as  those 
of  Si  and  52,  and  hence  the  lines  of  maximum  stress  shown  in 
Fig.  109  apply  equally  to  both.  At  the  upper  and  lower  surfaces 
of  the  beam  where  the  shear  is  zero,  the  unit-stress  5,  computed 
from  the  flexure  formula,  is  also  the  true  unit-stress ;  at  the  neutral 
surface  where  the  shear  is  the  greatest,  the  true  normal  stresses 
on  planes  where  there  is  no  shear  are  greater  than  the  apparent 
ones.  Since  the  unit-stresses  on  the  upper  and  lower  surfaces 
are  greater  than  for  any  other  points  in  a  cross-section,  it  is  never 
necessary  in  practical  problems  to  investigate  the  true  stresses 
in  a  beam. 

The  upper  surface  of  a  simple  beam  is  in  compression  while 
the  lower  surface  is  in  tension.  The  width  of  the  beam  hence 
suffers  a  lateral  expansion  in  its  upper  part  and  a  lateral  contrac- 
tion in  its  lower  part,  so  that  a  rectangular  section  becomes  a 
trapezoidal  one  when  the  load  is  applied.  This  change  is  so  slight 
that  it  is  rarely  observed,  but  there  is  little  doubt  that  it  can 
be  detected  by  precise  measurement.  For  example,  take  a  steel 
beam,  6x6  inches  in  section  and  so  loaded  that  the  flexural 
unit-stress  at  the  dangerous  section  is  30  ooo  pounds  per  square 
inch.  The  unit-shortening  of  the  upper  surface  and  the  unit- 
elongation  of  the  lower  surface  will  then  be  e  =  30  000/30  ooo  ooo 

=  0.001 ;    and  hence  the  total  lateral  contraction  of  the  width 
of  the  beam  at  the  dangerous  section  will  be  £  =  JXo.ooiX6 

=  0.002   inches,   a  quantity  that  can   be  easily  measured   with 
precise  calipers. 

A  uniform  load  resting  upon  the  upper  surface  of  a  simple 


368  TRUE  INTERNAL  STRESSES  CHAP.  XV 

beam  produces  a  vertical  compression  which  is  to  be  combined 
with  the  horizontal  compressive  unit-stress  in  order  to  obtain 
the  true  stresses.  Let  Si  be  the  flexurai  unit-stress  and  $2  the 
compressive  unit-stress  due  to  the  uniform  load.  Then  the 
true  maximum  compressive  unit-stress  in  a  horizontal  direction 
is  TI  =  Si  —  XS2)  while  that  in  a  vertical  direction  is  T2=S2  —  ASi. 
It  thus  appears  that  each  compression  diminishes  the  effect  of 
the  other.  Usually  S2  will  be  small  compared  with  Si,  so  that 
computations  are  rarely  necessary. 

A  concentrated  load  resting  upon  the  upper  surface  of  a 
simple  beam  may,  however,  produce  a  high  unit-stress  52.  The 
experiments  made  by  J.  B.  Johnson  in  1893,  on  the  contact 
between  the  surface  of  a  car  wheel  and  a  railroad  rail,  showed 
that  the  mean  compressive  unit-stress  was  about  80  ooo  pounds 
per  square  inch.  A  heavy  pressure  like  this  entirely  alters  the 
distribution  of  the  stresses  and  the  directions  of  the  lines  of  maxi- 
mum stress  in  its  vicinity,  for  TI  may  become  tension,  if  the 
elongation  due  it  can  occur.  In  the  contact  of  wheels  on  rails, 
however,  there  is  no  permanent  deformation  due  to  the  heavy 
vertical  compressive  stress,  which  indicates  that  lateral  flow  or 
elongation  could  not  occur.  Under  such  circumstances  there  is 
doubt  as  to  the  correctness  of  the  applicability  of  the  preceding 
theory  to  the  determination  of  the  true  stresses.  The  case  is 
perhaps  analogous  to  that  of  stresses  due  to  change  in  tempera- 
ture, where  heavy  stresses  may  arise  with  but  little  change  in 
length;  thus,  a  fall  of  200  degrees  Fahrenheit  in  temperature 
will  produce  a  unit-shortening  of  about  0.0014,  but  this  is  prob- 
ably sufficient  to  break  a  wrought-iron  bar,  if  it  is  prevented 
from  shortening  and  is  under  an  initial  tension  of  about  30  ooo 
pounds  per  square  inch. 

Prob.  1420.  A  steel  I  beam,  20  inches  deep  and  weighing  80  pounds 
per  linear  foot,  carries  a  uniform  load  of  24  ooo  pounds  on  a  span  of 
30  feet.  Compute  the  values  of  Si  and  S2  at  the  dangerous  section 
and  find  the  true  stresses. 

Prob.  1426.  How  must  a  simple  beam  be  loaded  so  that  the  elastic 
curve  is  an  arc  of  a  circle? 


ART.  143  STRESSES    DUE    TO    SHEAR  369 


ART.  143.     STRESSES  DUE  TO  SHEAR 

It  is  shown  in  Art.  6,  and  also  in  Art.  105,  that  forces  of  te^ 
sion  or  compression  acting  upon  a  body  produce  not  only 
internal  tensile  or  compressive  stresses,  but  also  internal  shear- 
ing stresses.  Conversely,  an  external  shear  acting  upon  a  body 
produces  in  it  not  only  internal  shearing  stresses,  but  also  internal 
tensile  and  compressive  stresses. 

For  example,  the  rectangle  A  BCD  in  the  web  of  a  plate 
girder,  shown  in  Fig.  143a,  may  be  considered.  Let  V  be  the 
shear  at  the  sections  AB  and  CD,  which 
are  taken  very  near  together  so  that  the 
weight  in  the  rectangle  itself  can  be  dis- 
regarded. This  vertical  shear  or  couple 
must,  be  accompanied  by  a  horizontal 
shear  FI,  which  in  this  case  is  caused  by 
the  resistance  of  the  flange  rivets.  Let 
the  thickness  of  the  material  be  one  unit; 
then  if  S  and  Si  are  the  shearing  unit- 
stresses,  their  values  are  S=V/AB  and  Fig.  143a 
S\  =  Vi/AD.  Now  taking  either  A  .or  D  as  an  axis  of  moments, 
the  equation  of  moments  is  VxAD  =  ViXAB,  and  hence  V/AB  = 
Vi/AD,  that  is,  the  shearing  unit-stresses  S  and  Si  are  equal. 
This  is  without  regard  to  the  weight  of  the  rectangle  itself,  which 
will  cause  a  slight  modification,  because  the  V  on  the  left  will 
then  be  greater  than  the  V  on  the  right.  But  if  AD  is  very 
small,  the  conclusion  is  strictly  true  that  the  horizontal  shearing 
unit-stress  is  equal  to  the  vertical  shearing  unit-stress. 

The  vertical  and  horizontal  shears  in  the  above  figure  tend 
to  deform  the  rectangle  into  a  rhomboid,  thus  causing  tension 
along  the  diagonal  BD  and  compression  along  the  diagonal  AC. 
At  every  point  in  the  rectangle,  then,  the  vertical  shearing  unit- 
stress  S  and  the  equal  horizontal  unit-stress  Si  combine  to  cause 
the  tension  and  the  compression  2*5  acting  with  inclina- 
tions of  45  degrees  to  the  shears.  Dividing  each  of  these  by 
the  area  2*  normal  to  its  direction,  it  is  seen  that  both  the  ten- 


370  TRUE  INTERNAL  STRESSES  CHAP.  XV 

sile  and  the  compressive  unit-stress  is  S;  that  is,  a  shearing 
unit-stress  causes  equal  tensile  and  compressive  unit-stresses  in 
directions  making  angles  of  45  degrees  with  the  shears. 

This  may  also  be  proved  from  the  discussion  in  Art.  105 
or  from  Art.  141.  Thus,  in  formula  (141)  let  the  axial  tensile 
unit-stress  S  be  made  zero,  then  the  maximum  tensile  and  com- 
pressive unit-stresses  Si  and  6*2  are  each  equal  to  S8.  If,  how- 
ever, Sa=o,  then  the  maximum  shearing  unit-stress  is  JS. 
Accordingly  an  axial  tension  or  compression  on  a  bar  produces 
a  shearing  unit-stress  equal  to  one-half  the  tensile  or  compres- 
sive unit-stress,  but  the  action  of  a  shear  produces  tensile  and 
compressive  unit-stresses  equal  to  the  shearing  unit-stress  itself. 
This  may  be  regarded  as  a  most  fortunate  arrangement  in  view 
of  the  fact  that  the  shearing  strength  of  materials  is  usually  less 
than  the  tensile  strength. 

The  above  relates  to  apparent  stresses  only.  The  true  stresses 
TI  and  T2,  corresponding  to  Si  and  S2,  are  those  that  correspond 
to  the  actual  deformations,  and  by  (139)  their 
values  are  Ti=Si  — 15*2  and  7^2  =^2  —  ASi, 
where  Si  and  S2  are  to  be  taken  as  positive 
for  tension  and  as  negative  for  compression. 
For  example,  let  Fig.  1436  represent  one  face  of 
a  cube  which  is  subject  to  the  shearing  unit- 
stress  S  of  5  ooo  pounds  per  square  inch,  each 
edge  of  the  cube  being  unity.  The  distortion 
of  the  square  is  shown  greatly  exaggerated  by  the  broken  lines, 
and  both  the  tension  along  the  longer  diagonal  and  the  com- 
pression along  the  shorter  diagonal  are  equal  to  5  ooo  pounds 
per  square  inch.  Now  if  the  factor  of  lateral  contraction  A  is 
J,  then  the  true  stress  along  the  longer  diagonal  is  TI  =  +6  250, 
while  that  along  the  shorter  one  is  T2=—  6250  pounds  per 
square  inch,  so  that  the  true  stresses  of  tension  and  compres- 
sion are  25  percent  greater  than  the  apparent  ones. 

It  may  be  noted  that  while  shear  produces  distortions,  it  does 
not  cause  changes  in  the  volume  of  a  body.  Thus,  for  the  above 
figure^  let  e  be  the  elongation  or  shortening  of  the  diagonals, 


ART.  144  TRUE    STRESSES    IN   SHAFTS  371 

then  the  length  of  the  longer  diagonal  is  2^  +  e  and  that  of  the 
shorter  diagonal  is  2*  — e;  the  area  of  the  rhombus  then  is 
J(2*  +  £)  (2*  -  e)  =  i,  which  is  the  same  as  that  of  the  square  before 
it  was  subjected  to  shear,  e2  being  a  negligible  quantity. 

Prob.  143.  A  steel  beam,  2X2X6  inches,  is  supported  at  its  ends 
and  has  a  concentrated  lead  40  ooo  pounds  at  its  middle.  Compute 
by  Art.  108  the  greatest  shearing  stress  which  occurs  at  the  neutral 
axis,  and  then  find  the  true  tensile  and  compressive  unit-stresses  which 
exist  there.  Draw  a  diagram  showing  the  directions  of  these  stresses. 

ART.  144.     TRUE  STRESSES  IN  SHAFTS 

When  a  round  shaft  is  acted  upon  by  torsion  alone,  the  stresses 
are  those  of  shearing,  and  these  act  along  every  section  normal  to 
the  axis,  the  maximum  S8  occurring  at  the  surface  (Art.  90). 
Any  square  in  one  of  these  normal  sections  is  hence  acted  upon 
by  two  equal  and  opposite  shears,  as  shown  in  Fig.  1436,  and  these 
produce  apparent  stresses  of  tension  and  compression  in  directions 
bisecting  those  of  the  shears.  The  discussion  of  the  last  article 
applies  in  all  respects  to  this  case,  and  from  it  the  true  stresses 
of  tension  and  compression  are  seen  to  be  each  equal  to  (i  +  X)Sa. 

When  a  horizontal  shaft  carries  a  load,  flexural  stresses  come 
into  action  and  these  must  be  combined  with  the  shearing  stresses 
in  the  manner  explained  in  Art.  106.  The  formulas  (141)  give 
the  apparent  and  formulas  (141)'  give  the  true  unit-stresses  due 
to  the  combination  of  torsion  and  flexure;  in  these  S  is  to  be  first 
computed  from  the  flexure  formula  (41),  while  S8  is  to  be  com- 
puted from  the  torsion  formula  (90)  or  from  the  special  formulas 
of  Art.  92.  From  the  last  paragraph  of  Art.  143,  it  is  to  be  con- 
cluded that  there  is  no  change  in  volume  of  the  shaft  under  torsion 
alone;  the  same  is  closely  the  case  when  flexure  is  added  to  the 
torsion,  because  the  decrease  in  volume  due  to  the  tension  is 
practically  the  same  as  the  increase  due  to  the  compression 
(Art.  13). 

The  compression  on  the  upper  surface  of  a  shaft  due  to  a 
load,  or  that  on  the  lower  surface  due  to  the  upward  reaction 


372  TRUE  INTERNAL  STRESSES  CHAP.  XV 

of  a  bearing,  produces  stresses  .which  act  normally  to  the  flexural 
stresses  of  tension  and  compression,  while  they  are  also  at  right 
angles  to  the  shearing  stresses  due  to  the  transmitted  torsion. 
A  formula  for  discussing  this  and  other  more  difficult  cases  is 
deduced  in  Art.  177,  and  an  application  of  it  to  the  above  case 
will  now  be  given.  Let  Sx  be  the  horizontal  flexural  unit- 
stress  at  the  surface  of  the  shaft,  Sy  the  vertical  compressive  unit- 
stress  due  to  the  load  or  bearing,  and  S8  the  shearing  unit-stress 
due  to  the  transmitted  torsion.  Then  in  formula  (177)  the 
value  of  A  is  Sx  +  Sy)  that  of  B  is  SxSy  —  S82,  and  that  of  C  is 
zero,  and  it  reduces  to  the  form 

S2  -  (Sx  +  SV)S+  SxSy  -Ss2  =0 

in  which  the  two  values  of  S  are  the  maximum  tensile  and  com- 
pressive unit-stresses;  solving  the  quadratic  equation  there  re- 

sults, 

*  (144) 


where  the  value  of  S  found  by  using  the  plus  sign  before  the 
radical  will  be  tension  or  compression  according  as  the  value 
of  %(Sx  +  Sy)  is  tension  or  compression.  When  either  Sx  or  Sy 
is  zero,  these  values  are  the  same  as  those  given  by  (141). 

As  a  numerical  example,  let  the  flexural  compressive  unit- 
stress  Sx  under  a  load  or  in  a  bearing  of  a  horizontal  steel  shaft 
be  3  ooo,  the  vertical  compressive  unit-stress  Sy  due  to  the  load 
or  bearing  be  i  200,  and  the  shearing  unit-stress  55  due  to  the 
torsion  be  6  ooo,  all  in  pounds  per  square  inch  Formula  (144) 
then  gives  £1  =  8200  pounds  per  square  inch  compression,  and 
S2  =  4  ooo  pounds  per  square  inch  tension  for  the  maximum 
normal  stresses;  also  S'  =  J(Si  —  £2)  =6  100  pounds  per  square 
inch  is  the  maximum  shearing  stress.  Lastly,  by  (139)  and  (140), 
the  corresponding  true  stresses  are  for  compression  7^  =  9500, 
for  tension  T2  =  6  700,  and  for  shear  T'  =  8  100  pounds  per  square 
inch.  In  common  practice  it  will  be  considered  that  the  greatest 
compression  is  3  ooo  and  the  greatest  shear  is  6  ooo  pounds  per 
square  inch,  but  the  result  of  this  investigation  shows  that  the 
true  compression  is  more  than  three  times  as  great  and  the  true 
shear  about  40  percent  greater. 


ART.  145  PURE   STRESSES  373 

Prob.  144.  Show  that  the  two  roots  of  (144)  are  always  real  whatever 
may  be  the  values  of  Sx,  Sy,  and  S8-  What  are  these  roots  when  Ss  is  o  ? 

ART.  145.    PURE  STRESSES 

The  term  '  pure  stress '  is  employed  for  cases  where  only 
one  kind  of  stress  exists.  When  a  plane  is  acted  upon  only  by 
forces  normal  to  it,  the  stress  on  the  plane  is  either  tensile  or  com- 
pressive  and  this  is  sometimes  called  'pure  normal  stress '.  When 
a  plane  is  acted  upon  only  by  forces  parallel  to  it,  the  stress  on 
the  plane  is  that  of  shearing,  and  this  is  sometimes  called  'pure 
shearing  stress '.  Upon  most  planes  in  the  interior  of  a  stressed 
body,  there  act  both  normal  and  shearing  stresses.  The  preceding 
articles  show  how  to  find  the  maximum  unit-stresses  Si  and  £2 
which  act  normally  against  certain  planes  upon  which  there  are 
no  shearing  stresses. 

Another  use  of  the  term  'pure  stress  '  is  with  respect  to  any 
and  all  planes  that  can  be  imagined  to  be  drawn  in  the  interior 
of  a  body.  When  the  forces  acting  upon  the  body  have  such 
values  that  there  can  be  no  shearing  stresses  within  it,  the  case 
is  called  one  of  'pure  internal  normal  stress'.  Referring  to 
Fig.  139  and  to  the  reasoning  in  the  last  paragraph  of  Art.  140, 
it  is  seen  that  there  can  be  no  shearing  stresses  on  any  planes 
within  the  body,  when  Si=S2  =  Ss;  this  is  the  case  where  the  unit- 
stresses  acting  on  the  six  faces  of  a  parallelepiped  are  all  equal. 
The  same  result  follows  when  a  body  is  acted  upon  by  equal 
compressive  forces  in  all  directions,  as  occurs  under  hydrostatic 
pressure.  Under  no  other  circumstances  can  the  interior  of  a 
stressed  body  be  free  from  shearing  stress,  and  hence  this  is  the 
only  case  of  pure  internal  normal  stress. 

There  is  no  case  of  a  body  having  only  internal  shearing  stress, 
for  the  discussion  of  Art.  143  shows  that  internal  shear  must 
always  be  accompanied  by  normal  stresses  which  act  in  directions 
bisecting  those  of  the  two  conjugate  shears.  There  may,  however, 
be  certain  planes  within  a  body  upon  which  only  shearing  stresses 
act.  In  order  to  find  such  planes,  let  Figs.  140#  and  I40b  be  again 
considered,  and  let  the  forces  shown  in  the  latter  be  resolved 


374 


TRUE  INTERNAL  STRESSES 


CHAP.  XV 


normal  to  the  diagonal  dz.  Let  S  be  the  normal  unit-stress  of 
tension  or  compression  on  dz;  then  the  total  stress  on  that  diagonal 
is  Sdz  and  this  is  equal  to  Sidy  .  sin/9  +  S2dx.  cos/9.  Replacing 
dx/dz  and  dy/dz  by  their  values  cos/9  and  sin#,  there  is  found 
S  =  Si  sin2#  +  S2cos26.  Now  when  S  =  o,  there  is  no  normal 
stress  on  the  plane  that  makes  the  angle  6  with  the.  direction  of 
Si',  this  occurs  when  tan#  =  (  —  6*2/51  )*  and  on  the  plane  thus 
determined  only  shearing  stresses  are  acting.  It  is  seen  that 
no  value  of  0  is  possible  unless  Si  and  S2  have  contrary  signs, 
that  is,  one  must  be  tension  and  the  other  compression.  When 
S2=-±Si,  then  tan#=±o-5  and  #=±26J°;  when  S2=-Si, 
then  tan 6  =±i  and  0  =  ± 45° ;  when  S2  =  - 3-Si ,  then  tan/9  =±1.73 
and  0=  ±60°,  and  so  on.  Hence  for  each  negative  value  of  S2/Si 
there  are  two  planes  equally  inclined  to  the  direction  of  Si  upon 
which  only  shearing  stresses  act.  The  following  figures  show 
the  three  cases  computed  above. 


if- 


The  shearing  unit-stresses  on  these  planes  of  true  shear  are 
not  as  great  as  those  on  the  planes  bisecting  the  directions  of 
Si  and  52,  for  on  the  latter  the  maximum  shears  exist.  For 
Fig.  1456,  however,  where  the  normal  tension  and  compression 
are  numerically  equal,  the  planes  of  pure  shear  coincide  with 
those  of  maximum  shear;  this  is  the  case  most  frequently  men- 
tioned as  one  of  pure  shear  (Fig.  1456),  but  the  above  investiga- 
tion shows  that  there  may  be  many  other  cases. 

The  term  '  pure  flexure  '  is  used  for  a  part  of  a  beam  where 
there  are  no  vertical  shears.  For  instance,  take  a  simple  beam 
and  subject  it  to  two  concentrated  loads,  each  equal  to  P  and 
placed  at  equal  distances  from  the  supports.  Then  there  is  no 
vertical  shear  between  the  loads,  and  hence  the  flexural  stresses 
above  the  neutral  surface  are  pure  compression,  while  those 


ART.  146  INTERNAL  FRICTION  375 

below  it  are  pure  tension.  In  testing  a  beam  it  is  sometimes 
advantageous  to  subject  it  to  two  equal  concentrated  loads  placed 
at  equal  distances  from  the  middle;  thus  the  bending  moment 
between  the  loads  is  constant,  and  the  changes  of  length  of 
the  fibers  are  uniform  at  equal  distances  from  the  neutral 
surface.  The  experiments  of  Talbot  on  steel-concrete  beams, 
described  in  Art.  116,  were  made  in  this  way.  It  must  be  noted, 
however,  that  there  prevail  in  all  directions,  except  horizontally 
and  vertically,  shearing  unit-stresses  accompanying  the  pure 
tension  and  compression;  if  S  is  the  flexural  unit-stress  at  any 
point  between  the  loads,  then  JS  is  the  maximum  shearing  unit- 
stress  which  makes  angles  of  ±45°  with  the  direction  of  S. 

Prob.  1450.  In  the  formula  Sn=Si  sm26+S2  cos20,let  Si  be  larger 
than  $2-  Show  that  the  values  of  Sn  cannot  be  greater  than  Si  nor 
less  than  6*2.  This  is  the  equation  of  a  curve  in  polar  coordinates, 
Sn  being  the  radius  vector  for  the  variable  angle  0;  what  kind  of  a 
curve  is  it? 

Prob.  1456.  A  parallelepiped  is  acted  upon  by  normal  unit-stresses 
of  6  400  and  2  800  pounds  per  square  inch  in  directions  at  right  angles 
to  each  other,  the  first  being  tension  and  the  second  compression. 
Compute  the  pure  shearing  unit-stress  and  the  maximum  shearing 
v\nit-stress,  and  find  their  directions. 


ART.  146.    INTERNAL  FRICTION 

In  all  the  preceding  discussions,  the  applied  forces  and 
the  internal  stresses  have  been  supposed  to  be  in  equilibrium, 
this  being  the  case  where  the  applied  forces  have  attained  the 
full  magnitudes  so  that  no  further  deformation  of  the  body  occurs. 
Other  considerations  enter  during  the  period  while  the  defor- 
mations are  occurring  under  applied  forces  which  increase  from 
zero  up  to  their  final  values.  During  this  period  there  are  motions 
of  the  molecules,  and  this  motion  is  resisted  by  internal  friction, 
just  as  the  motion  of  a  book  upon  a  table  is  opposed  by  the  fric- 
tion between  the  surfaces  of  contact.  The  planes  of  maximum 
stress,  found  in  the  preceding  articles,  are  hence  not  the  cor- 


376  TRUE  INTERNAL  STRESSES  CHAP.  XV 

reel  planes  of  greatest  stress  during  the  period  while  the  defor- 
mation of  a  body  is  occurring. 

The  subject  of  internal  friction  was  first  recognized  in  the 
experiments  made  by  Tresca,  about  1860,  on  the  flow  of  metals 
under  high  compressive  stress,  but  it  was  not  until  after  1890 
that  it  received  careful  attention.  In  1893  the  remarkable  dis- 
covery was  made  by  Hartmann  that  lines  of  stress  became  visible 
on  the  surface  of  polished  metallic  specimens  when  the  elastic 
limit  of  the  material  was  reached  or  surpassed,  and  that  these 
lines  remained  after  the  loads  were  removed.  Fig.  146a  repre- 
sents such  lines  for  specimens  of  rectangular  section,  and  it  is 
seen  that  in  compression  they  make  an  angle  with  the  axis  less 
than  45  degrees,  while  in  tension  this  angle  is  greater  than  45 
degrees.  It  was  observed  that  the  sum  of  these  two  angles  was 
always  90  degrees  for  the  same  metal  and  that  the  directions 
of  the  lines  were  independent  of  the  size  and  length  of  the  speci- 
men and  of  the  unit-stress.  The  number  of  lines,  however, 
increased  as  the  unit-stress  increased  from  the  elastic  limit  to 
the  ultimate  strength.  For  the  case  of  tension,  Hartmann  found 
that  the  angle  (f>  which  the  lines  make  with  the  axis  of  the  bar 
was  65  degrees  for  nickel  steel,  63  degrees  for  tempered  steel, 
and  58  degrees  for  annealed  steel;  for  compression  the  angle  0 
between  the  lines  and  the  axis  was  the  complement  of  <£. 

When  round  specimens  of  metal  with  polished  surfaces  were 
subjected  to  stresses  above  the  elastic  limit  of  the  material,  it. 
was  found  that  the  lines  were  not  straight  but  spiral,  as  shown 
in  Fig.  1466,  these  spirals  making  the  same  angle  with  the  axis 
as  the  straight  lines  on  the  rectangular  specimens.  Under  the 
microscope  it  was  noted  that,  in  general,  these  lines  were  depres- 
sions below  the  intermediate  surfaces  and  that  the  larger  lines 
seen  by  the  naked  eye  were  really  several  small  lines  very  near 
together.  Hartmann  also  experimented  on  spheres  and  beams, 
finding  that  curved  lines  appeared  on  their  polished  surfaces 
when  the  elastic  limit  of  the  material  was  reached  or  surpassed, 
their  directions  always  remaining  the  same  in  the  same  speci- 
men. The  lines  for  a  beam  shown  in  his  book  Deformation* 


ART.  146 


INTERNAL  FRICTION 


377 


dans  les  Metaux  (Paris,  1896)  have  very  little  resemblance  to 
any  of  the  theoretic  lines  of  maximum  stress  which  are  shown 
in  Fig.  109. 

These  interesting  lines  probably  indicate  the  directions  of  the 
planes  or  surfaces  on  which  the  sliding  or  shearing  of  the  mate- 
rial is  beginning  to  occur.  This  supposition  is  strengthened  by 
the  phenomena  of  the  rupture  of  brittle  materials  under  com- 
pression, where  it  is  found  that  the  failure  ultimately  takes  place 
by  shearing  along  planes  inclined  to  the  axis  of  the  specimen, 
as  Fig.  1696  shows  for  cement  and  timber.  These  planes  make 
angles  with  the  axis  varying  from  10  to  40  degrees,  the  angle  for 
stone  usually  being  about  20  and  that  of  cement  and  concrete 
about  30  degrees.  Also  it  is  observed  that  metallic  bars  under 
tension  sometimes  rupture  with  an  oblique  or  cup-like  fracture, 
the  inclination  of  which  to  the  axis  is  50  degrees  or  more.  It  may 
therefore  be  regarded  as  almost  demonstrated  that  materials  begin 
to  fail,  both  in  tension  and  compression,  by  shearing  along  oblique 
planes,  and  that  the  commencement  of  the  failure  is  at  the  time 
the  elastic  limit  of  the  material  is  reached. 


t  t   t   t 


t  t  t 


Fig.  146a 


Fig.  1466 


The  theory  of  internal  stress,  set  forth  in  the  preceding  articles 
of  this  chapter,  shows  that  the  maximum  shearing  unit-stresses, 
both  apparent  and  true,  are  those  upon  planes  making  angles  of 
45°  with  the  axis  of  the  bar.  Since  the  actual  planes  of  failure 
<ire  greater  than  45°  for  tension  and  less  for  compression,  it  must 
be  concluded  that  some  resisting  force  acts  during  the  progress 


378  TRUE  INTERNAL  STRESSES  CHAP.  XV 

of  the  deformation  which  has  not  heretofore  been  considered, 
and  this  resistance  is  probably  that  of  friction.  Much  attention 
has  been  given  to  this  question  since  1895,  and  the  work  of  Rejto, 
published  in  1897,  endeavors  to  account  for  the  brittleness,  plas- 
ticity, ductility,  and  even  the  strength  of  materials,  by  the  help 
of  coefficients  of  internal  friction. 

Prob.  146.  Consult  Rejto's  Innere  Reibung  der  festen  Korper, 
and  explain  his  formula  for  the  tensile  strength  of  materials. 

ART.  147.     THEORY  OF  INTERNAL  FRICTION 

When  one  surface  begins  to  slide  on  another,  the  ratio  of  the 
force  parallel  to  the  sliding  surface  to  the  normal  pressure  is  called 
the  coefficient  of  friction;  it  is  an  abstract  number  and  may 
be  designated  by  v.  Let  N  be  the  normal  pressure  or  stress 
between  the  two  bodies  and  F  the  force  which  just  begins  to  cause 
motion  when  it  acts  parallel  to  the  surface  of  contact,  then  the 
approximate  law  of  sliding  friction  is  given  by  F  =  vN.  This 
law  may  be  applied,  tentatively  at  least,  to  the  case  of  a  bar 
under  axial  stress,  during  the  period  while  the  stress  is  increasing 
up  to  its  final  value,  and  it  may  be  supposed  that  sliding  or  shearing 
is  then  beginning  to  occur  along  surfaces  indicated  by  the  lines 
and  planes  described  in  the  last  article.  There  is  hence  a  coeffi- 
cient of  internal  friction  /*,  which  is  not  necessarily  the  same  as 
that  of  sliding  friction  but  which  may  be  used  in  the  same  manner 
by  means  of  the  law  F  =  j^N. 

The  simplest  case  is  that  of  a  bar  subject  to  axial  compression, 
as  in  Fig.  147 a.  Let  the  section  area  be  unity  and  S  be  the 
compressive  unit-load  which  causes  the  equal  axial  unit-stress 
S1  on  all  planes  normal  to  the  axis.  Let  any  plane  mn  be  drawn 
cutting  this  bar,  and  let  0  be  the  angle  which  it  makes  with  the 
axis.  When  there  is  no  axial  stress  in  the  bar,  there  exists  normal 
to  this  plane  a  molecular  unit-force  So  which  binds  together  the 
two  parts  so  that  no  motion  can  occur.  When  the  axial  com- 
pressive unit-stress  is  applied,  this  causes  a  compressive  stress 
S  sin#  normal  to  the  plane,  and  since  the  area  of  the  plane  is 


ART.  147  THEORY  OF   INTERNAL   FRICTION  379 

i/sin0,  the  compressive  unit-stress  is  S  sin2#.  The  total  normal 
pressure  per  square  unit  on  the  plane  then  is  N  =  So  -f  S  sin2#. 
The  force  acting  parallel  to  the  plane  is  S  cos#,  or  F  =  S  cos#  sin0 
for  each  square  unit.  Accordingly,  if  motion  is  just  beginning, 

Scosdsind=n(S0+Ssm2fl)        or        S=/*S0/(sin0  cos0-//sin20) 

Now  let  6  be  regarded  as  varying  from  o  to  90°,  then  the  unit-load 
S  required  to  cause  motion  will  vary  from  <x>  to  SQ,  and  it  follows 
that  the  motion  will  actually  begin  along  that  plane  which  has 
such  a  value  of  6  that  S  shall  be  a  minimum;  or  the  value  of  5 
which  causes  motion  to  begin  is  less  for  the  plane  of  motion  than 
for  any  other  plane.  This  requires  that  sin#  cos#  —  /i  sin2#  shall 
be  a  maximum,  and  the  value  of  6  which  renders  it  such  is  found 
to  be  given  by  cot  26  =  /*.  Hence, 

(147) 


gives  the  relations  between  the  quantities  ft,  6,  and  SQ.  The 
observed  angle  6  is  always  less  than  45°  ;  for  different  qualities  of 
steel,  6  lies  between  25°  and  32°  and  hence  the  coefficient  of 
internal  friction  lies  between  0.84  and  0.49. 


^Nr 


Fig.  147a  Fig.  1476 

In  considering  the  case  of  tension  shown  in  Fig.  1476,  let  (/> 
be  the  angle  which  a  plane  mn  makes  with  the  axis.  Then  the 
normal  unit-pressure  on  the  plane  is  SQ  —  S  sin2</>,  and  the  force 
per  unit  of  area  acting  parallel  to  the  plane  is  S  sin^>  cos<£.  For 
the  case  of  incipient  motion,  the  law  of  friction  then  gives, 

S  sin2<  or          S  =  fiS0/  sin<> 


and,  by  the  same  reasoning  as  before,  it  is  concluded  that  the 
actual  plane  of  motion  is  that  which  renders  sin<£  cos<£  +  /£  sin2<£ 
a  maximum,  and  this  maximum  obtains  when  cot2<£=-//. 
Accordingly, 


-cot  2<j>          ftSo  =  %S  tan  <£  =  J5(/£  +V'i  +  /<2)        (147)' 


380  TRUE  INTERNAL  STRESSES  CHAP.  XV 

are  the  equations  applicable  to  tension  in  which  $  is  observed  to 
be  always  greater  than  45  degrees. 

Since  the  coefficient  of  internal  friction  jj.  must  be  regarded 
as  a  constant  for  the  same  material,  it  follows  that  the  angles  Q 
and  (j>  are  necessarily  complementary,  for  by  equating  the  two 
values  of  n  the  relation  between  the  angles  is  given  by  #  +  </>  =90°. 
The  above  theory  may  hence  be  said  to  explain  why  it  is  that 
the  lines  described  in  the  last  article  make  an  angle  with  the  axis 
which  is  less  than  45  degrees  for  compression  and  greater  than 
45  degrees  for  tension,  and  why  these  angles  are  comple- 
mentary. 

It  seems  a  reasonable  assumption  to  regard  /iSo  as  the  ulti- 
mate shearing  strength  Ss  of  the  material,  since  j*S0  equals  the 
force  per  unit  of  area  which  will  cause  shearing  along  the  plane. 
When  a  brittle  specimen  is  ruptured  by  direct  compression, 
failure  generally  occurs  by  shearing  along  one  or  more  planes 
which  make  an  angle  6  with  the  axis  less  than  45  degrees,  am. 
accordingly  Ss  =  JSctan#  may  be  written  as  a  tentative  formula, 
for  the  relation  between  the  ultimate  shearing  strengths  S8  and 
the  ultimate  compressive  strength  Sc.  The  following  are  rough 
approximate  values  of  6  as  observed  in  compressive  tests,  together 
with  the  values  of  /*  and  S8/SC  as  computed  from  (147)  : 

Anthracite  Coal  #  =  15°  /*  =  I-73  S8=o.~LT)Sc 

Sandstone  6  =  20°  jj.  =  i.ig  Ss=o.i8Sc 

Hard  Brick  #  =  25°  ^=0.84  Ss=o.2^Sc 

Cement  and  Concrete  #  =  30°  ^=0.58  Ss  =  o.2gSc 

Cast  Iron  #  =  35°  ^  =  0.36  5«= 


These  computations  indicate  that  the  coefficient  of  internal 
friction  is  the  highest  and  that  the  ratio  of  the  shearing  to  the  com- 
pressive strength  is  the  lowest  for  the  most  brittle  material.  Thus 
for  cast  iron  the  shearing  strength  is  35  percent  of  the  compres- 
sive strength,  according  to  this  computation,  while  for  anthra- 
cite coal  it  is  only  13  percent. 

When  two  bars  of  steel  are  stressed  up  to  their  elastic  limits, 
one  in  compression  and  the  other  in  tension,  the  elastic  limit 


ART.  147  THEORY  OF  INTERNAL   FRICTION  381 

is  closely  the  same  for  both  bars,  and  it  hence  seems  that  S  in 
(147)  should  be  the  same  as  S  in  (147)',  which  requires  6  and 
(j>  to  be  equal.  This  is  a  result  altogether  at  variance  with  experi- 
ment, and  it  must  hence  be  concluded  either  that  S0  in  tension 
is  different  from  So  in  compression  or  that  the  above  reasoning 
is  defective  in  failing  to  include  one  or  more  elements  that  must 
ultimately  be  introduced  in  order  to  perfect  the  theory.  Much 
work  still  remains  to  be  done  on  this  important  subject,  both 
in  theory  and  by  experiment,  before  definite  ideas  can  be  formed 
regarding  the  true  internal  stresses  which  prevail  while  a  body 
is  undergoing  deformations.  The  theory  of  Arts.  139-145  relates 
only  to  static  stresses,  namely,  to  those  which  occur  when  the 
applied  forces  have  attained  their  full  magnitudes,  so  that  both 
external  and  internal  equilibrium  prevails.  This  static  theory 
appears  to  be  correct  in  every  detail  for  static  stresses  which 
do  not  surpass  the  elastic  limit  of  the  material,  but  the  perma- 
nency of  the  lines  of  shearing  seen  upon  polished  metallic  speci- 
mens, seems  to  throw  a  doubt  upon  its  entire  applicability  to 
cases  where  the  elastic  limit  is  surpassed,  even  though  com- 
plete equilibrium  exists.  As  far  as  true  internal  static  stresses 
are  concerned,  this  theory  is  indeed  not  necessarily  valid,  for 
formulas  (139)  apply  only  within  the  elastic  limit,  but  for  the 
apparent  static  stresses  it  should  be  valid  for  all  cases.  In  order 
that  the  full  and  complete  truth  may  be  ascertained,  further 
studies  on  internal  friction  and  on  internal  molecular  forces  are 
absolutely  necessary. 

In  conclusion  it  may  be  noted  that  the  idea  of  internal  fric- 
tion throws  light  upon  the  fatigue  of  materials  under  repeated 
stresses  (Art.  137).  For  the  case  of  compression  where  heat 
is  evolved  for  stresses  both  below  and  above  the  elastic  limit, 
it  is  not  difficult  to  see  that  energy  is  expended  in  changing  the 
positions  of  the  molecules  at  each  application  of  stress;  for  the 
case  of  tension  the  same  occurs  for  stresses  higher  than  about 
three-fifths  of  the  elastic  limit.  The  material  is  hence  fatigued 
or  changed  in  structure  by  the  internal  friction,  and  this  change 
should  be  greater  for  large  ranges  of  stress  than  for  small  ones. 
Undoubtedly  the  complete  explanation  of  fatigue  is  closely  allied 


382  TRUE  INTERNAL  STRESSES  CHAP.  XV 

to  that  of  internal  friction  and  to  changes  in  internal  molecular 
forces.  The  indications  are  that  any  stress,  however  small,  will 
produce  fatigue  when  it  is  repeated  a  number  of  times  in  a  material 
that  has  a  crystalline  structure.  Steel  is  such  a  material  and  the 
various  crystals  that  are  seen  in  it  under  the  microscope  have 
cleavage  planes  which  are  weakened  by  the  detrusion  due  to  re- 
peated stress. 

Prob.  147#.  Five  constants  have  thus  far  been  used  in  this  volume  as 
applicable  to  a  material  when  stressed  up  to  its  elastic  limit  in  tension. 
What  are  these  constants? 

Prob.  1476.  Consult  The  lion  and  Steel  Magazine  for  July,  1905, 
and  read  an  article  on  the  failure  of  an  iron  plate  through  fatigue. 
Also  consult  other  volumes  of  this  periodical,  and  ascertain  the  names 
and  characteristics  of  the  various  crystals  which  are  seen  in  steel. 


A.RT.  148  FACTS  AND  PRINCIPLES  383 


CHAPTER  XVI 
GUNS  AND  THICK  CYLINDERS 

ART.  148.     FACTS  AND  PRINCIPLES 

THE  discussion  of  pipes  under  internal  pressure,  given  in 
Art. '30,  was  made  under  the  assumption  that  the  thickness  of 
the  pipe  is  small  compared  to  its  diameter,  so  that  the  tensile 
stress  of  the  metal  of  the  pipe  might  be  regarded  as  uniformly 
distributed.  Pipes  under  high  internal  pressure  have  thicknesses 
sometimes  nearly  equal  to  the  inside  diameter,  and  in  such  the 
tensile  stresses  are  not  uniformly  distributed.  The  steel  pipe 
used  to  transmit  water  pressure  to  the  large  forging  press  of  the 
Bethlehem  Steel  Company,  in  South  Bethlehem,  Pa.,  has  an 
inside  diameter  of  16  inches,  a  thickness  of  8  inches,  and  it  is 
subject  at  times  to  a  pressure  of  5  600  pounds  per  square  inch. 
The  theory  of  the  investigation  and  design  of  thick  pipes  will  be 
developed  in  this  chapter. 

The  guns  used  in  modern  warfare  are  subject  to  high  internal 
pressures  produced  by  the  explosion  of  the  powder.  These 
pressures  are  measured  by  noting  their  effects  in  shortening  small 
copper  cylinders  and  comparing  these  deformations  with  those 
produced  by  known  loads  in  a  testing  machine.  In  this  manner 
it  has  been  ascertained  that  powder  pressures  of  50  ooo  pounds 
per  square  inch  are  often  produced  during  the  firing  of  a  gun, 
while  the  extreme  pressure  of  88  ooo  pounds  per  square  inch  has 
been  observed  in  a  special  experiment.  In  order  that  the  metal 
of  the  gun  may  not  be  stressed  beyond  its  elastic  limit  linger 
these  heavy  pressures,  it  is  necessary  that  the  thickness  should 
be  large.  For  inside  diameters  greater  than  3  inches,  the  gun 
is  generally  formed  of  two  or  more  concentric  cylinders,  the  inner 
one  being  called  the  c  tube '  and  the  others  *  hoops ' ;  these  hoops 
are  shrunk  upon  the  tube  (Art.  32)  so  that  they  produce  com- 
pression in  it  and  thus  enable  it  to  carry  a  heavier  powder  pressure 


384  GUNS  AND  THICK  CYLINDERS  CHAP.  XVI 

than  a  solid  tube  of  the  same  total  thickness.     The  first  hoop 
around  the  gun  tube  is  often  called  the  'jacket'. 

Fig.  148  gives  a  longitudinal  and  a  cross  section  of  a  gun  having 
two  hoops,  the  breech  block  that  closes  the  powder  chamber  A 
being  omitted.  This  breech  block  can  be  swung  open  to  admit 
the  projectile  and  the  powder  case,  the  position  of  the  former 
being  at  B.  Before  the  explosion  the  breech  block  is  swung  into 
place  and  locked.  At  the  instant  of  the  explosion  the  pressure 
of  the  gas  is  the  greatest  since  it  is  then  confined  to  the  spaces  A 
and  B',  this  part  of  the  gun  is  called  the  breech,  and  here  it  is 
that  the  greatest  thickness  is  required ;  over  the  breech  and  extend- 
ing some  distance  forward,  the  figure  shows  two  hoops  E  and  F 
surrounding  the  gun  tube  D.  As  the  projectile  moves  toward 
the  muzzle,  the  gas  occupies  a  larger  volume,  so  that  its  pressure 
decreases  and  becomes  zero  as  the  projectile  leaves  the  gun. 
The  tube  C  has  spiral  grooves  cut  in  its  inside  surface,  which  cause 
the  projectile  to  have  a  rotary  motion.  In  designing  a  gun  of 
this  kind  it  is  required  that  the  section  areas  at  the  forward  end 
of  each  hoop  shall  be  sufficient  to  resist  the  maximum  powder 
pressure  which  can  there  be  exerted. 


IA 


F 

Fig.  148 

Modern  guns  are  made  of  hard  steel,  often  of  fluid  compressed 
steel,  which  has  an  elastic  limit  of  about  50  ooo  and  an  ultimate 
strength  of  about  90  ooo  pounds  per  square  inch.  The  allowable 
working  stresses  are  large,  in  some  cases  as  large  as  the  elastic 
limit.  Most  careful  workmanship  and  rigid  inspection  are 
exercised  in  their  manufacture,  and  constant  improvements  are 
made  in  their  design.  These  modern  guns  have  been  entirely 
developed  since  1870,  prior  to  which  time  cast-iron  cannon  were 
mainly  in  use.  They  are  usually  designed  for  a  powder  pressure 
of  50  ooo  pounds  per  square  inch. 


ART.  149 


LAME'S  FORMULA 


385 


Prob.  148.  The  diameter  of  a  powdei  chamber  is  8f  inches  and 
that  of  the  projectile  is  8  inches,  while  the  pressure  during  the  explosion 
is  48  ooo  pounds  per  square  inch.  Compute  the  total  forward  pres- 
sure on  the  projectile  and  the  total  backward  pressure  on  the  breech 
block.  Where  does  the  difference  of  these  pressures  take  effect? 


ART.  149.    LAME'S  FORMULA 

Let  a  thick  hollow  cylinder,  shown  in  Fig.  1490,  be  subject  to 
a  pressure  RI  on  each  square  unit  of  the  inside  surface  and  to  a 
pressure  ^2  on  each  square  unit  of  the  outside  surface.  The 
inside  pressure  may  be  produced  by  the  expansion  of  a  gas  and 
the  outer  pressure  by  the  atmosphere  or  by  other  causes.  It  is 
required  to  determine  the  internal  stresses  produced  by  these 
pressures  at  any  point  in  the  cylindrical  annulus  not  very  near 
the  end.  The  length  of  the  cylinder  is  to  be  regarded  as  con- 
siderably longer  than  the  outer  diameter,  in  order  that  the  dis- 
turbing influence  of  the  ends  may  not  affect  the  reasoning. 


t 

t   4 

T                                      I 

1 

Fig.  149a 

Let  r\  and  r2  be  the  inside  and  outside  radii;  then  the  inside 
pressure  on  the  end  of  the  closed  cylinder  is  nri2R\  and  the  outside 
pressure  on  that  end  is  nr22R2.  The  usual  case  is  that  where  the 
inside  is  greater  than  the  outside  pressure,  and  then  n(ri2Ri—r22R2) 
is  the  longitudinal  tension  in  the  annulus.  For  any  part  of  the 
cylinder,  not  very  near  the  end,  this  tension  must  be  uniformly 
distributed  over  the  cross-section  of  the  annulus.  The  longi- 
tudinal tensile  unit-stress  So  in  the  annulus  is  hence  a  constant 
for  all  points,  and  its  value  is  found  by  dividing  the  total  tension 
by  the  section  area  a;  whence, 


This  longitudinal  unit-stress,  together  with  the  radial  pressures, 


386  GUNS  AND  THICK  CYLINDERS  CHAP.  XVI 

causes  a  longitudinal  elongation  of  the  cylinder,  which  is  also 
to  be  regarded  as  uniform  for  all  parts  of  the  annulus,  not  very 
near  the  end.  Under  these  assumptions,  the  theory  of  true  stress 
given  in  Art.  139  may  be  applied  to  the  determination  of  the  unit- 
stresses  at  any  point  in  the  cylindrical  annulus. 

Let  x  be  the  distance  from  the  axis  of  the  cylinder  to  any 
point  in  a  cross-section   of  the  cylindrical  annulus.      Any  ele- 
mentary particle  is  here  held  in  equilibrium 
by  the  longitudinal  unit-stress  SQ,  a  tan- 
gential  unit-stress   S,    and    a    radial   unit- 
stress  R.     The  value  of  R  is  evidently  inter- 
~r~_i  mediate  between  RI  and  R2;    in  Fig.  1496 

both    R   and    S   are    regarded    as    tensile. 
Flg'  l  Now   from   Art.  139  the   effective   longitu- 

dinal unit -elongation  of  the  cylinder  due  to  these  three  stresses  is- 

£0  =  T/E  =  (S0  -  /IS  -  IR)/E 

in  which  A  is  the  factor  of  lateral  contraction  the  mean  valu'e 
of  which  for  wrought  iron  and  steel  is  about  J.  But,  as  above 
noted,  both  £0  and  S0  are  constant  for  all  parts  of  the  annulus, 
and  it  hence  follows  that, 

S+R  =  constant  or 


which  is  one  equation  between  the  unit-stresses  S  and  R  when 
Ci  is  a  quantity  whose  value  is  to  be  determined  by  establishing 
a  second  equation. 

Let  an  elementary  annulus  of  thickness  dx  be  drawn;  its 
inner  radius  is  x  and  its  outer  radius  is  x  +  dx.  The  pressure 
for  one  unit  of  length  in  a  direction  perpendicular  to  any  diam- 
eter is  Rx  for  the  inner  surface  and  (R  +  dR)(x+dx)  for  the  outer 
surface  of  this  elementary  annulus.  Thus,  exactly  as  in  the 
case  of  a  thin  pipe  (Art.  30),  the  equation  of  equilibrium  between 
acting  pressure  and  resisting  stress  is, 

(R+dR)(x+dx)-Rx=Sdx  or  xdR+Rdx=Sdx 

which  is  a  second  equation  between  the  unit-stresses  S  and  R. 

The  solution  of  these  two  equations  is  readily  made  by  insert- 
ing 2C\  —  R  for  S  in  the  second  equations  and  integrating;  then, 


ART.  149  LAME'S  FORMULA  387 


where  €2  is  a  constant  of  the  integration  the  value  of  which  is 
to  be  determined  by  regarding  the  limiting  values  of  R,  these 
being  the  inner  and  outer  unit-pressures  RI  and  R2.  It  is  best 
to  regard  these  unit-pressures  as  without  sign,  and  then  R=  —  RI 
when  x  =  ri,  and  R=—R2  when  x  =  r2',  inserting  these  condi- 
tions in  the  second  formula,  there  result  two  equations  from  which, 

Ci  =  (ri^  -rfR2)/(rf-r?)  C2  =  /vV22(tfi  ~R2)/(r22-r12) 

and,  inserting  these  constants  in  (149),  are  now  obtained, 
S=  (r^R,  -r22R2+r-^(Rl  -R2)\  I  (r22  -  rft 

2   2  (149)' 

R  =  (r?Ri  -  r22R2  -^  (R,  -R2)j  /  (r22  -n2) 

which  are  Lame's  formulas  for  the  tangential  and  radial  unit- 
stresses  in  hollow  cylinders  under  inside  and  outside  pressures. 
In  deriving  these  formulas,  both  S  and  R  have  been  supposed 
to  be  tension;  this  will  be  the  case  if  their  values  are  positive, 
while  a  negative  value  will  indicate  compression. 

The  tangential  unit-stress  S  is  usually  greater  than  the  radial 
unit-stress  R,  and  is  the  controlling  factor  in  the  design  of  guns 
and  thick  cylinders.  It  is  seen  to  increase  as  x  decreases,  and 
hence  it  is  the  greatest  at  the  inside  surface  of  the  cylinder;  it 
may  be  either  tension  or  compression,  depending  upon  the  rela- 
tive values  of  the  given  radii  and  unit-pressures  The  radial 
unit-stress  R  is  always  compression,  its  value  ranging  from  RI 
at  the  inside  surface  to  R2  at  the  outside  surface, 

As  a  numerical  example,  let  a  cylinder  be  one  foot  in  inside 
and  two  feet  in  outside  diameter,  the  inside  pressure  being  600 
and  the  outside  15  pounds  per  square  inch.  Here  ri=6,  r2  =  i2, 
^i  =6co,  ^2  =  15,  and  the  formulas  become, 

5  =  180+28  oSo/x2  #  =  180-28  o8o/*2 

Here  x  varies  between  6  and  12  inches,  and  S  ranges  from  +960 
to.  +375  pounds  per  square  inch,  while  R  ranges  from  —600 
to  -15  pounds  per  square  inch,  +  denoting  tension  and  —  denot- 


388 


GUNS  AND  THICK  CYLINDERS 


CHAP.  XVI 


ing  compression.  The  tangential  unit-stress  S  is  hence  about  2j 
times  greater  at  the  inside  surface  of  the  hollow  cylinder  than 
at  the  outside  surface. 

Prob.  149.  A  solid  cylinder  is  subject  to  a  uniform  radial  pressure 
of  14  ooo  pounds  per  square  inch  on  its  surface.  Prove  that  the  radial 
unit-stress  is  uniform  throughout  the  cylinder.  Prove  that  the  tangen- 
tial unit-stress  is  uniform  throughout  the  cylinder,  and  find  its  value. 

ART.  150.     THICK  PIPES  AND  SOLID  GUNS 

Lame's  formulas  have  been  widely  used  for  the  discussion 
of  water  pipes  and  solid  guns.  The  inside  unit-pressure  R\  is 
usually  large  compared  to  the  pressure  of  the  atmosphere  on 
the  outside  surface  so  that  the  latter  may  be  neglected.  The 
unit-stresses  S  and  R  at  any  point  within  the  cylindrical  annulus 
then  are, 


A  discussion  of  this  equation  shows  that  the  tangential  unit-stress 
S  is  greatest  when  x  =  r\  and  least  when  x  =  r2,  and  that  the 
radial  unit-stress  R  varies  similarly.  Thus,  for  the  inside  surface, 

Si=Ri(r22+rf)/(rf-rf}  and  R=-R^          (150) 

while  for  the  outside  surface  of  the  cylinder, 

S2==Ri  '  27*i  7(^2  — Y\  )  and  R=o 

and   accordingly  the   greatest  unit-stresses,   as  given   by   (150), 

are  those  generally  used  in  investigation  and  design.  The  man- 
ner in  which  5  varies 
throughout  the  annulus 
is  shown  in  Fig.  150  by 
the  arrows.  It  is  seen 
B  that  the  different  parts 
of  the  annulus  are  un- 
equally stressed  under 
the  tangential  tension, 
the  law  of  variation  of 
the  unit-stresses  being 

that  which  is  expressed  by  the  first  equation  in  formula  (149). 


A.RT.  150  THICK  PIPES  AND  SOLID  GUNS  389 

As  a  special  case,  let  the  radius  of  the  outside  surface  be 
double  that  of  the  inside  surface,  or  r2  =  2^1.  Then  for  the  inside 
surface  where  x  =  r±  the  tangential  unit-stress  is  5i=f^i;  for 
the  outside  surface,  where  JG  =  r2,  it  is  £2  =  §^2 ;  for  a  point  half 
way  between  these  surfaces  it  is  f  f/?i,  and  all  of  these  are  ten- 
sion. For  the  same  case  the  radial  unit-stress  for  the  inside 
surface  is  —  Ri,  for  the  outside  surface  it  is  o,  and  for  a  point 
half  way  between  them  it  is  —  J-Ri,  all  being  compression  as 
indicated  by  the  minus  sign. 

The  first  formula  of  (150)  was  formerly  much  used  for  the 
investigation  of  guns  and  thick  pipes,  and  it  is  still  valuable  for 
general  discussions.  As  an  example,  let  a  solid  steel  gun  have  an 
inside  diameter  of  7.5  inches  at  the  powder  chamber  and  the 
thickness  of  the  tube  be  1.75  inches.  Let  it  be  required  to  find 
the  greatest  tensile  unit-stress  produced  when  the  inside  pres- 
sure from  the  explosion  is  10  ooo  pounds  per  square  inch.  Here 
ri=3-75  inches,  ^2  =  5.50  inches,  ^1  =  10000  pounds  per  square 
inch.  Then  from  the  formula  Si  is  found  to  be  27  300  pounds 
per  square  inch,  which  is  but  little  more  than  one-half  the  elastic 
limit  of  gun-steel,  and  hence  the  degree  of  security  is  ample. 

As  an  example  of  design,  let  the  inside  diameter  be  3.25 
inches,  the  pressure  caused  by  the  explosion  15  ooo  pounds  per 
square  inch,  and  the  allowable  unit-stress  in  tension  be  30  ooo 
pounds  per  square  inch ;  and  let  it  be  required  to  find  the  outside 
diameter.  Here  7-1  =  1.625  inches,  ^1  =  15000,  and  ,$1=30000 
pounds  per  square  inch.  Then  solving  for  r2)  there  results, 

r2  =  ri[(S1  +  Rl)/(S1-Rl)^  (150)' 

from  which  the  outside  radius  r2  is  found  to  be  2.815  inches; 
thus  the  thickness  of  the  tube  is  1.19  inches,  and  its  outside 
diameter  is  5.63  inches. 

When  formula  (150)  is  applied  to  a  thin  pipe,  r2  is  to  be  re- 
placed by  r\  +  /,  where  /  is  the  thickness,  and  t2  may  be  neglected 
when  it  is  to  be  added  to  r{*.  The  formula  then  reduces  to 
tSi=Ri(r  +  t),  which  is  slightly  more  accurate  than  that  of  (30), 
and  it  gives  slightly  larger  values  of  Si  in  investigation  and  slightly 
larger  thicknesses  in  design.  Neglecting  /  in  comparison  with 


390  GUNS  AND  THICK  CYLINDERS  CHAP.  XVi 


r,  this  becomes  tSi  =  riR,  which  is  the  same  as  the  common  for- 
mula for  thin  pipes  derived  in  Art.  30.  Either  of  these  formulas, 
however,  would  lead  to  grave  error  when  applied  to  a  pipe  whose 
thickness  is  as  great  as  one-half  its  diameter. 

From  formula  (150)'  it  is  seen  that  r2  becomes  infinite  when 
RI  equals  S,  that  is,  the  inside  unit-pressure  must  never  be  greater 
than  the  allowable  tensile  unit-stress  of  the  material.  Cast-iron 
cylinders  for  small  forging  presses  have  been  used  under  pressures 
as  high  as  5  ooo  pounds  per  square  inch,  and  it  hence  follows 
that  the  actual  unit-stress  5  must  have  been  much  higher  than 
this;  the  thickness  of  such  cylinders  is  usually  greater  than  the 
inside  radius.  The  indications  of  experience  are  that  factors  of 
safety  for  thick  pipes  under  pressure  may  be  much  lower  than 
for  thin  ones. 

Prob.  150.  A  solid  gun  tube  is  6  inches  in  diameter  and  3  inches 
thick.  What  is  the  inside  pressure  that  will  produce  a  maximum  tan- 
gential tension  of  30  ooo  pounds  per  square  inch  ? 

ART.  151.     A  COMPOUND  CYLINDER 

In  a  solid  gun  the  maximum  tension  occurs  at  the  inside  sur- 
face during  the  explosion,  rising  suddenly  from  o  up  to  its  greatest 
value  Si.  If  now  the  metal  near  the  bore  can  be  brought  into 
compression,  this  initial  stress  must  be  overcome  before  the 
tension  can  take  effect,  and  thus  the  capacity  to  resist  the  inside 
pressure  is  increased.  One  method  of  producing  this  compression 
is  by  means  of  a  hoop,  or  jacket,  shrunk  upon  a  tube  so  as  to 
produce  an  outside  unit-pressure  R2  over  the  surface  of  the  tube. 
This  arrangement  may  be  called  a  hollow  compound  cylinder. 

In  its  normal  state  of  rest,  the  inner  cylinder  or  tube  has  no 
pressure  on  its  inner  surface  and  R2  on  each  square  unit  of  its 
outside  surface.  Making  RI=O  in  (149)'  and  also  x  =  r\  and 
x=r2  in  succession,  there  are  found 

Si  =  -R2  -  2r22/(rJ  _fl2)  s2=  -R2  .  (r<?  +  n*)/(rf-n*) 

which  are  the  tangential  unit-stresses  at  the  inside  and  outside 
surfaces  of  the  tube  due  to  the  external  pressure  ^2-  Both  of 


ART.  151  A  COMPOUND  CYLINDER  391 

these  are  compression,  but  the  former  is  numerically  greater 
than  the  latter,  since  ir22  is  greater  than  r22  +  ri2.  If  the  hoop 
is  to  be  shrunk  on  so  as  to  produce  a  compressive  unit-stress  Se 
at  the  inner  surface  of  the  tube,  the  unit-pressure  R2  upon 
the  outer  surface  must  be, 

R2=-Se(r22-r^/2r22 

and  the  shrinkage  may  be  so  regulated  as  to  produce  this  pressure 
R2  in  the  normal  state  of  rest.  Then  the  tangential  stresses 
throughout  the  tube  are  all  compression,  while  the  radial  pressures 
range  from  R2  on  the  outer  surface  to  o  on  the  inner  surface. 

As  an  example,  let  ^  =  2  inches,  7*2  =  3  inches,  and  let  it  be 
required  to  find  the  outer  pressure  which  will  cause  a  tangential 
compressive  unit-stress  of  18  ooo  pounds  per  square  inch  at  the 
inside  surface  of  the  tube.  Here  the  last  formula  gives  R2 
=  5  ooo'  pounds  per  square  inch,  and  hence  the  hoop  must  be 
shrunk  upon  the  tube  so  as  to  produce  this  radial  pressure  at  the 
surface  of  contact  of  hoop  and  tube. 

When  the  gun  is  fired,  the  explosion  of  the  powder  causes 
an  internal  tangential  tension  S  given  by  (149)',  the  greatest 
value  of  which  is  at  the  inside  surface  of  the  tube.  Making  x=r\f 
this  tensile  unit-stress  is  found  to  be, 

Si  =[(ri2+r22)R,  -  2r22R2]/(r2*-rl2) 


which  is  Lame's  formula  for  the  investigation  of  the  tube  of  a 
compound  gun.  The  first  term  of  the  second  member,  namely 
+  (ri2  +  r22)Ri/(r22-ri2),  is  the  tensile  unit-stress  at  the  bore  in 
case  there  is  no  hoop,  while  the  second  term  —  2r22R2/(r22  —  ri2) 
is  the  compressive  unit-stress  due  to  the  shrinkage  of  the  hoop. 
In  Fig.  155  the  line  aai  represents  the  first  term,  a  A  represents 
the  second  term,  and  A  ai  represents  the  resultant  tensile  unit- 
stress  6*1  during  firing.  This  formula  shows  that  if  R2  be  made 
equal  to  (ri2  +  T22)Ri/2r2,  then  Si  will  become  zero,  but  so  great 
a  radial  hoop  compression  is  never  used  in  the  design  of  com- 
pound guns. 

For  example,  let  a  tube  whose  inside  and  outside  diameters 
are  4  and  6  inches  be  hooped  so  that  a  radial  compression  of 


392  GUNS  AND  THICK  CYLINDERS  CHAP,  xvi 

5  ooo  pounds  per  square  inch  is  exerted  at  the  common  surface 
of  tube  and  hoop,  and  let  the  inside  pressure  due  to  the  powder 
explosion  be  25  ooo  pounds  per  square  inch.  It  is  required  to  find 
the  resultant  tangential  unit-stress  at  the  bore  during  the  firing. 
Here  r2  =  3  and  7-1  =  2  inches,  ^2  =  5000  and  ^1  =  25000  pounds 
per  square  inch.  Then,  from  the  formula,  the  resultant  tension 
at  the  bore  during  the  firing  is  found  to  be  +65  000—18000  = 
+  47  ooo  pounds  per  square  inch.  If  this  tube  has  no  hoop  the 
tangential  tension  at  the  bore  is  65  ooo  pounds  per  square  inch; 
hence  the  very  great  advantage  of  the  hoop  in  diminishing  the 
stress  at  the  bore  during  firing  is  apparent.  As  another  example 
let  the  data  be  the  same,  except  that  ^2  =  10000  pounds  per 
square  inch;  then  +65  ooo  — 36000= +29  ooo  pounds  per 
square  inch  is  the  tangential  unit-stress  at  the  bore. 

Prob.  151.  A  gun  tube  3  inches  in  diameter  and  1.5  inches  thick 
is  hooped  so  that  the  tangential  compression  on  the  inside  surface 
is  30  ooo  pounds  per  square  inch.  What  powder  pressure  RI  will 
produce  a  resultant  tangential  tension  on  the  inside  surface  of  30  ooo 
pounds  per  square  inch? 


ART.  152.     CLAVARINO'S  FORMULAS 

The  preceding  method  of  investigating  gun  tubes  is  defec- 
tive in  that  the  two  unit-stresses  S  and  R  are  the  apparent  and 
not  the  true  internal  unit-stresses.  It  was  shown  in  Art.  139  that 
the  true  stresses  are  those  corresponding  to  the  actual  deforma- 
tions, and  that  they  are  determined  from  the  apparent  stresses 
by  help  of  the  factor  of  lateral  contraction  L  For  gun-steel  the 
value  of  A  is  usually  taken  as  J.  Lame's  formulas  were  deduced 
in  1833,  but  it  was  not  until  about  1880  that  they  were  modified 
by  Clavarino  so  as  to  give  the  true  internal  stresses. 

At  any  point  in  the  annulus  of  a  hollow  cylinder  (Fig.  1496) 
the  apparent  tangential,  radial,  and  longitudinal  unit-stresses  are 
S,  R,  and  S0.  Let  T  be  the  true  tangential  unit-stress,  then 
from  (139)  its  value  is  T  =  S-XR-XS0',  inserting  in  this  the 
values  of  S,  R,  and  So  found  in  Art.  149,and  taking  the  factor 


ART.  152  CLAVARINO'S   FORMULAS  393 

of  lateral  contraction  as  J,  it  reduces  to, 

(r2^r^  (152) 


which  is  Clavarino's  formula  for  the  tangential  unit-stress.  This 
is  the  principal  formula  for  the  investigation  of  steel  guns  and 
thick  pipes. 

This  formula  shows,  as  before,  that  the  tangential  stress  is 
greatest  at  the  inside  surface  of  the  cylinder.  Making  X  =  TI,  the 
true  internal  maximum  unit-stress  is  found  to  be, 

7\  =[(n2+  4r22)Ri  ~  Sr2*R2]/3  (r22  -  r?)  (152)' 

which  is  the  practical  formula  for  the  discussion  of  the  most 
common  cases.  T\  may  be  either  tension  or  compression,  depend- 
ing upon  the  relative  values  of  the  pressures  and  radii. 

Clavarino's  formulas  have  been  generally  used  since  1885  in 
the  investigation  and  design  of  guns,  instead  of  those  of  Lame. 
In  order  to  compare  them,  the  particular  case  where  the  outer 
diameter  is  double  the  inner  diameter  may  be  considered.  Here 
r2  =  2ri  and  formulas  (149)'  and  (152)'  reduce  to, 

Si  =4(5*i  -8*2)  T,  =1(17*!  -20*2) 

Now  if  R2=o,  the  first  formula  gives  a  smaller  unit-stress  than 
the  second;  if  R2=Ri,  the  first  gives  a  unit-stress  three  times 
as  large  as  the  second;  if  RI=O,  the  first  gives  a  value  some- 
what larger  than  the  second.  Thus,  since  the  second  formula 
gives  undoubtedly  a  better  representation  of  the  true  stress  than 
the  first,  it  follows  that  Lame's  method  errs  on  the  side  of  danger 
for  a  solid  gun  and  on  the  side  of  safety  for  a  hooped  tube.  The 
value  J  here  used  for  the  factor  of  lateral  contraction  is  that 
employed  in  the  United  States  by  both  the  Army  and  the  Navy 
in  gun  formulas,  and  also  generally  in  Europe;  in  France,  how- 
ever, the  value  e  =  J  is  adopted. 

For  a  thick  pipe  or  solid  gun,  the  outside  pressure  R2  is  zero, 
and  formula  (152)r  can  then  be  written  in  the  forms, 


the  first  of  which  may  be  used  for  the  investigation  and  the  second 


394  GUNS  AND  THICK  CYLINDERS  CHAP.  XVI 

for  the  design  of  hollow  cylinders.  According  to  the  second 
formula,  the  inside  unit-pressure  RI  can  never  be  as  large  as  three- 
fourths  of  the  allowable  tensile  unit-stress  TI,  and  this  may  be 
taken  higher  than  for  a  thin  pipe.  For  example,  let  it  be  required 
to  find  the  thickness  of  a  steel  pipe  8  inches  in  inside  diameter, 
under  a  head  of  water  of  i  200  feet,  this  including  the  effect  of 
water  ram  (Treatise  on  Hydraulics,  Art.  148),  the  allowable 
tensile  unit-stress  to  be  15000  pounds  per  square  inch.  Here 
Ti=i5ooo,  and  ^1=0.434X1200  =  520  pounds  per  square 
inch,  and  then  the  formula  gives  7-2=4.18  inches,  whence  the 
required  thickness  is  0.18  inches.  From  the  old  formula  (150)', 
the  thickness  is  found  to  be  0.22  inches. 

Prob.  152.  Solve  Pioblem  150  by  the  formulas  of  this  article  and 
compare  the  results  fo  ind  by  the  two  methods. 


ART.  153.     BIRNIE'S  FORMULAS 

The  preceding  articles  present  an  outline  of  the  methods  of 
investigating  stresses  in  guns  by  the  formulas  of  Lame  and 
Clavarino.  The  formulas  of  Lame  refer  to  apparent  stresses 
only ;  those  of  Clavarino,  although  referring  to  true  stresses, 
are  not  strictly  correct  for  hooped  guns  at  rest  because  they  are 
deduced  for  a  hollow  cylinder  with  closed  ends.  Now  a  gun 
at  rest  has  no  inside  pressure  and  no  closed  end  and  hence 
there  can  be  no  external  longitudinal  stress  upon  it;  accordingly 
the  unit-stress  So  should  be  zero  for  a  gun  at  rest.  The  true 
tangential  unit-stress  T  then  is  S  —  ^R;  using  the  values  of  S 
and  R  deduced  in  Art.  149,  there  results  for  any  point  in  the 
annulus,  at  a  distance  x  from  the  axis, 

1-R2)}/3(r22-r12)         (153) 

which  is  Birnie's  principal  formula  for  the  discussion  of  hooped 
guns  at  rest.     Making  x  =  r\,  this  becomes, 

TI  =[(2ri2+4r22)#  i  -6r22R2]/3(r22-ri2)  (153)' 

which  is  the  tangential  unit-stress  at  the  inside  surface  of  a  hoop 


.  153  BIRNIE'S  FORMULAS  395 

with  the  radii  r\  and  r%  and  is  under  the  pressures  R\  and  R2. 
For  the  gun  tube  itself,  RI  is  o  during  the  state  of  rest. 

Birnie's  formulas  are  used  in  the  ordnance  bureau  of  the 
United  States  Army  for  the  discussion  of  gun  tubes  and  hoops, 
both  at  rest  and  during  the  firing.  To  compare  the  formulas  of 
Clavarino  with  those  of  Birnie,  the  particular  case  of  a  hoop 
or  pipe  where  r2  =  i.2r\  may  be  considered.  Then  (152)'  and 
(153)'  reduce  to, 

TI  =  5.  i2Ri  -5.  56^2  and  TI  =  5.88^1  -6.35^2 

Now  if  R2  =o,  as  for  a  solid  gun  during  firing,  the  second  formula 
gives  a  tangential  unit-stress  15  percent  larger  than  the  first; 
if  R\=o,  as  for  a  hooped  gun  at  rest,  the  second  gives  a  unit 
stress  1  8  percent  larger  than  the  first.  Thus  for  this  case  Cla- 
varino's  formulas  appear  to  err  toward  the  side  of  danger. 

Birnie's  formulas  apply  only  to  hoops  and  tubes  upon  which 
the  longitudinal  stress  is  zero,  and  this  is  not  the  case  during 
the  explosion.  For  a  solid  gun,  or  for  a  tube  attached  to  the 
breech  block,  a  more  correct  formula  may  be  found  by  con- 
sidering the  actual  value  of  S0  due  to  the  inside  pressure.  Here 
the  longitudinal  pressure  is  nr\2Ri,  and  this  produces  longitu- 
dinal tension  upon  the  area  x(r22  —  ri2),  so  that  So  =ri2Ri/(r£  —  fi2) 
is  the  apparent  longitudinal  unit-stress.  The  true  tangential 
stress  T  at  any  point  in  the  annulus  then  is  S  —  ^R  —  JSo> 
accordingly, 


This  gives  values  of  T  lower  than  those  found  from  the  formulas 
of  Clavarino  and  Birnie.  For  x  =  r\  it  becomes, 

TI  =[(ri2+4r22)R!  -6r22R2]/3(r22-r12) 

which  is  the  true  tangential  unit-stress  at  the  inside  surface  of 
a  hooped  gun  during  the  explosion,  R2  being  the  pressure  upon 
its  outside  surface  due  to  the  shrinkage  of  the  hoop.  For  a  simple 
gun  tube  or  water  pipe  where  R2  =  o,  this  formula  agrees  with 
that  of  Clavarino;  for  a  hooped  tube  at  rest,  it  gives  a  value  of 
TI  which  is  20  percent  greater  than  that  of  Clavarino. 


396  GUNS  AND  THICK  CYLINDERS  CHAP.  XVI 

Prob.  153.  Solve  Problem  150  by  the  formulas  of  this  article  and 
compare  the  results  with  those  of  Problem  152. 


ART.  154.     HOOP  SHRINKAGE 

Let  e  be  the  elongation  or  contraction  of  any  radius  x,  then. 
2ne  is  the  elongation  or  contraction  of  any  circumference  2xx. 
Now  27ie/2nx  is  the  change  in  the  circumference  per  unit  of 
length  due  to  the  unit-stress  T\  hence  e/x=T/E,  and  e  =  (T/E)x 
is  the  change  in  length  of  the  radius  of  the  circle  due  to  the  tan- 
gential unit-stress  T.  When  x  =  r\9  the  deformation  e\  is  that 
of  the  radius  of  the  bore  due  to  the  unit-stress  T\;  if  x  =  r2,  the 
deformation  e2  is  that  of  the  outside  radius  where  the  unit- 
stress  is  T2. 

Suppose  a  compound  cylinder  to  be  formed  by  shrinking  a 
hoop  upon  a  tube.  The  inside  radius  of  the  tube  is  r\  and  its 
outside  radius  r2 ;  the  inside  radius  of  the  hoop  is  r2  and  its  outside 
radius  r%.  In  consequence  of  the  shrinkage  the  radial  unit-pres- 
sure ^2  is  produced  between  the  two  surfaces;  this  causes  the 
tube  to  be  under  tangential  compression 
and  the  hoop  to  be  under  tangential  ten- 
sion. It  is  required  to  find  these  stresses 
when  the  original  inside  radius  of  the  hoop 
is  less  than  that  of  the  outside  radius  of  the 
tube  by  the  amount  e. 

Let  e2  be  the  decrease  in  the  outside 
radius  of  the  tube  and  e2  the  increase  in 
the  inside  radius  of  the  hoop;  then  e  =  e2  +  e2.  In  Fig.  154, 
which  is  much  exaggerated,  cd  represents  e2  and  be  represents 
e2.  Also,  let  T2  be  the  tangential  compression  at  the  outside 
surface  of  the  tube  due  to  the  shortening  e2,  and  let  T2  be  the 
tangential  tension  at  the  inside  surface  of  the  hoop  due  to  the 
elongation  e2.  Then 

e=(T2/E)r2+  (T2'/E)r2  or  T2+  T2'=Ee/r2 

which  gives  one  equation  between  T2  and  T2. 

Formula  (153)'  is  applied  to  the  tube  by  making  RI=O  and 


ART.  154  HOOP    SHRINKAGE  397 

x  =  r2',   thus  the  tangential  compression  is, 


Formula  (153)  is  applied  to  the  hoop  by  replacing  RI  by  R2)  R2 
by  o,  r\  by  r2,  and  r2  by  r3;  then  for  x  =  r2,  there  results, 

T2'  =  R2(2r22+4r32)/3(r32-r22)  =pR2 

which  is  the  tangential  tension.  Dividing  the  first  of  these  ex- 
pressions by  the  second,  there  is  found, 

T2/T2f=a/fi  or  fiT2=aT2' 

which  is  a  second  equation  between  T2  and  T2. 

The  solution  of  these  two  equations  furnishes  the  values  of 
T2  and  T2  in  terms  of  known  quantities;   then, 

r2=7^'^o  T2'=T'1^T3  (154) 

in  which  a.  and  ft  depend  only  on  the  radii,  or, 

and  thus  the  tangential  compression  at  the  outside  surface  of  the 
tube  and  the  tangential  tension  at  the  inside  surface  of  the  hoop 
may  be  computed.  The  tangential  compression  at  the  bore  is, 
however,  greater  than  T2,  and  it  may  be  found  from  (153)  by 
substituting  the  value  of  R2)  now  known,  and  making  oc  =  r\\ 
thus, 


is  the  greatest  compressive  unit-stress  in  the  tube  due  to  the 
radial  pressure  of  the  hoop. 

As  a  numerical  example,  let  a  compound  cylinder  be  formed 
of  a  steel  tube  whose  inside  radius  is  3  inches  and  outside  radius 
5  inches,  with  a  steel  hoop  whose  thickness  is  2  inches.  It  is 
required  to  find  the  stresses  produced  when  the  original  difference 
between  the  outside  radius  of  the  tube  and  the  inside  radius  of 
the  hoop  is  0.004  inches.  First,  the  sum  of  the  two  tangential 
stresses  at  the  surface  of  contact  is  (Ee)/r2  =  24  ooo  pounds  per 
square  inch,  if  E  is  taken  as  30  ooo  ooo  pounds  per  square  inch. 
Secondly,  from  the  given  radii,  the  value  of  a  is  found  to  be 
43/24,  and  that  of  /?  to  be  41/12.  Then  formulas  (154)  give 
T2  =  g  260  pounds  per  square  inch  at  the  outside  surface  of  the 


398  GUNS  AND  THICK  CYLINDERS  CHAP,  xvi 

tube  and  TV  =  15  74°  pounds  per  square  inch  for  the  inside  sur- 
face of  the  hoop.  Thus  it  is  seen  that  the  hoop  tension  is  nearly 
double  the  compression  on  the  outside  surface  of  the  tube.  At 
the  bore  of  the  tube,  however,  the  tangential  compression  is  found 
to  be  TI  =  14  400  pounds  per  square  inch. 

The  decrease  in  the  outside  radius  of  the  tube  is  next  com- 
puted and  found  to  be  £2  =  0.00138  inches,  while  the  increase 
in  the  inside  radius  of  the  hoop  is  ^2'  =  0.00262  inches.  Hence 
if  the  radius  of  the  common  surface  of  contact  is  to  be  exactly  5 
inches  after  the  shrinkage,  the  tube  should  be  turned  to  an  outside 
radius  of  5.0014  inches,  and  the  hoop  to  an  inside  radius  of  4.9974 
inches.  The  radius  of  the  bore,  however,  will  then  be  less  than 
3  inches  by  the  quantity  (T\r\/E)=  0.00144  inches;  hence  if  its 
final  diameter  is  to  be  exactly  6.0000  inches,  it  should  be  turned 
to  a  diameter  of  6.0029  inches. 

The  formula  of  Birnie  has  been  used  in  solving  the  above 
numerical  example;  if  that  of  Clavarino  is  used,  the  following 
values  will  be  found :  T2  =  7  030,  Tj  =  16  970,  TI  =  14  400  pounds 
per  square  inch;  e2  =  0.00117,  e2  =  0.00283,  and  ^1=  0.00144 
inches.  The  shrinkages  thus  agree  within  0.0002,  which  is  as 
close  as  measurements  can  be  relied  upon. 

The  above  investigation  closely  applies  to  the  case  of  a  hoop 
or  crank  web  shrunk  upon  a  solid  shaft  or  solid  crank  pin  (Art.  97) 
by  making  r\=o  and  letting  r2  be  the  mean  radius  of  the  web. 
For  example,  let  fi=o,  r2  =  i6  inches,  ^3  =  24  inches,  and  let  e/r2 
be  rsViF  in  accordance  with  the  old  rule  (Art.  32).  Then  a  =  §, 
/?=  fX4.4,  and  the  tangential  compression  at  the  outside  surface 
of  the  shaft  or  pin  is  T2  =  3  740  pounds  per  square  inch,  while 
the  tangential  tension  at  the  surface  of  the  hoop  or  web  is 
7Y  =  i6  500  pounds  per  square  inch;  the  radial  compression  in 
the  crank  or  pin  is  TI  =  1 1  300  pounds  per  square  inch.  It  thus 
appears  that  the  ratio  e/r2  =  ^-s\-$  gives  shrinkage  stresses  which 
are  higher  than  advisable  when  the  other  stresses  which  act  upon 
the  web  and  pin  are  considered  (Art.  98). 

Prob.  154.  A  solid  steel  shaft,  6  inches  in  radius,  is  to  be  hooped 
so  that  the  greatest  tensile  stress  in  the  hoop  and  the  greatest  com- 


ART.  155  DESIGN   OF   HOOPED    GUNS  399 

pressive  stress  in  the  shaft  shall  be  1 5  ooo  pounds  per  square  inch. 
Find  the  thickness  of  the  hoop  and  the  radius  to  which  its  inside 
surface  should  be  turned. 


ART.  155.     DESIGN  OF  HOOPED  GUNS 

A  hooped  gun  should  be  so  constructed  that  neither  the 
stresses  due  to  hoop  shrinkage  nor  those  developed  during  the 
firing  shall  exceed  the  elastic  limit  of  the  material.  The  simple 
case  of  a  tube  with  one  hoop  can  here  only  be  considered.  If 
the  radii  are  given,  as  also  the  inner  pressure  RI  due  to  the  ex- 
plosion, it  may  be  desired  to  find  the  shrinkages  so  that  this 
requirement  will  be  fulfilled.  As  J^i  is  very  large,  it  is  desirable 
that  the  given  dimensions  should  be  such  as  to  require  the  least 
amount  of  material. 

The  condition  of  minimum  amount  of  material  will  be  in 
general  fulfilled  when  the  stresses  during  the  explosion  are  as 
great  as  allowable  and  as  nearly  equal  as  possible.  The  diagram 
in  Fig.  155  represents  the  distribution  of  the  internal  stresses 
under  this  supposition.  O  is  the  center  of  the  gun,  OA  the 
inside  radius  r\,  while  AB  is  the  thickness  of  the  tube  and  BC 
that  of  the  hoop.  The  shaded  areas  show  the  stresses  due 
to  hoop  shrinkage,  A  a  and  Bb  being  the  tangential  compres- 
sions TI  and  T2  of  the  last  article,  while  Bb'  is  the  tangential 
tension  7Y,  and  Cc  is  the  tangential  tension  at  the  outer  sur- 
face of  the  hoop.  When  the  explosion  occurs  the  two  cylinders 
are  thrown  into  tangential  tension,  Aa\  and  Bbi  being  those 
at  the  inner  surfaces  of  the  tube  and  hoop.  The  above  prin- 
ciple indicates  that  both  Aa\  and  Bb i  should  be  equal  to  the 
maximum  allowable  unit-stress  Te,  which  for  guns  is  often  taken 
nearly  as  high  as  the  elastic  limit  of  the  material. 

In  designing  a  hooped  gun,  the  radius  of  the  bore  and  the 
thickness  of  the  tube  may  be  assumed,  and  it  may  be  required 
to  find  the  thickness  and  shrinkage  of  the  hoop  so  that  the  stresses 
Aa,  Aa-i,  and  Bbi  in  Fig.  155  are  each  equal  to  the  elastic  limit 
of  the  material.  Or,  given  the  radius  of  the  bore  and  the  out- 


400 


GUNS  AND  THICK  CYLINDERS 


CHAP.  XVI 


side  radius  of  the  hoop,  it  may  be  required  to  find  the  interme- 
diate radius  under  the  same  conditions.  These  problems  can 
be  solved,  as  well  as  more  complex  ones  relating  to  guns  with 
several  hoops.  Guns  with  seven  hoops  have  been  built,  but  the 
usual  number  is  three  or  four. 


Fig.  155 

The  formulas  of  Arts.  153  and  154  may  be  applied  to  the 
design  of  a  gun  by  assuming  the  allowable  tangential  unit-stresses, 
as  also  the  thicknesses  of  the  tube  and  hoops.  For  a  given  unit- 
pressure  RI  due  to  the  explosion,  the  shrinkages  are  then  to 
be  computed.  This  method  is  one  frequently  used,  and  it  will 
here  be  illustrated  for  a  gun  with  one  hoop.  Let  ^1=3.04,  r2  = 
5.80,  and  ^3  =  9.75  be  the  given  radii,  and  let  50000  pounds  per 
square  inch  be  the  allowable  unit-stress  for  both  tension  and 
compression.  It  is  required  to  find  the  radii  to  which  the  sur- 
faces shall  be  turned  so  that  their  values  shall  be  those  above 
given  when  the  gun  is  at  rest.  These  radii  will  be  readily  found 
when  the  tangential  unit-stresses  TI  and  T2  for  the  tube  and 
T'2  and  T$  for  the  hoop  have  been  computed,  since  these  deter- 
mine the  changes  in  length  of  the  radii.  The  first  step  is  to 
compute  the  numbers  o:  and  /?,  which  are  found  to  be  1.424  and 
2.425  respectively.  Since  T\  is  to  be  50  ooo  pounds  per  square 
inch  compression  for  the  inside  surface  of  the  tube,  T2  for  the 
outer  surface  will  be  T\(2r-p  +  r£')/$r£  =  2$  800  pounds  per 
square  inch  compression,  and  accordingly  for  the  common  sur- 
face of  tube  and  hoop  R2=  r2/a  =  i8  100  pounds  per  square 
inch.  For  the  inside  surface  of  the  hoop,  T2=^R2  =  ^  900 
pounds  per  square  inch  tension.  For  the  outside  surface  of  the 


ART.  156  DESIGN   OF  HOOPED    GUNS  401 

hoop  where  -#3=0,  formula  (153)  may  be  used  by  increasing 
each  of  the  subscripts  by  unity  and  making  oc  =  r3,  thus  giving 
T3  =  2R2r22/(r32-r22)  =  i9  800  pounds  per  square  inch  tension. 
Then  the  change  in  the  inside  radius  of  the  gun  tube  is  (Ti/E)ri  = 
0.0051  inches,  and  hence  the  bore  must  be  turned  to  a  radius  of 
3.0400  +  0.0051=3.0451  inches  in  order  that  it  may  be  exactly 
3.04  inches  after  the  hoop  is  shrunk  on;  the  change  in  the  out- 
side radius  is  (T2/E) r2  =  0.0050  inches,  so  that  the  outside  sur- 
face of  the  tube  must  be  turned  to  a  radius  of  5.800  +  0.0050  = 
5.8050  inches.  The  change  in  the  inside  radius  of  the  hoop 
is  (T2/E)r2  =  0.0085  inches,  so  that  its  inner  surface  must  be 
turned  to  a  radius  of  5.8000  —  0.0085  =  5.7915  inches;  the  change 
in  the  outside  radius  of  the  hoop  is  (Ts/E)r3=  0.0064  inches, 
so  that  its  outside  surface  must  be  turned  to  a  radius  of 
9.7500  —  0.0064  =  9.7436  inches. 

This  gun  must  now  be  investigated  to  find  what  powder 
pressure  will  cause  the  stresses  T\  and  T%  to  be  50  ooo  pounds 
per  square  inch  tension  during  the  explosion.  If  T'^  has  this 
value,  the  part  of  it  due  to  the  powder  explosion  is  50  ooo  —  43  900  = 
6  100  pounds  per  square  inch;  hence  the  radial  compression 
between  the  tube  and  the  hoop  which  is  due  to  the  explosion 
must  be  ^2  =  6  ioo//?=2  500  pounds  per  square  inch.  The 
value  of  TI  due  to  the  explosion  is  100  ooo  pounds  per  square 
inch  tension,  since  the  initial  compression  of  50  ooo  pounds  per 
square  inch  must  first  be  overcome.  Inserting  then  in  (153)' 
the  values  T\  =  100000,  ^2  =  2500,  ^12  =  9.242,  r22  =  33-640,  and 
solving  for  RI  gives  RI  =  $I  100  pounds  per  square  inch,  which  is 
the  highest  allowable  powder  pressure.  Under  this  pressure  the 
unit-stresses  represented  by  Aa,  Adi,  and  Bbi  in  Fig.  155  are 
each  50  ooo  pounds  per  square  inch,  while  all  other  tangential 
and  radial  stresses  have  smaller  values. 

In  conclusion  it  may  be  noted  that  this  chapter  has  been 
prepared  in  order  to  present  the  general  principles  of  the  design 
of  guns,  rather  than  to  give  the  detailed  methods  which  are 
followed  when  three  or  more  hoops  are  used.  The  work  of 
Meigs  and  Ingersoll  (Baltimore,  1885)  and  that  of  Story  (Fort 


402  GUNS  AND  THICK  CYLINDERS  CHAP.  XVI 

Monroe,  1894),  each  being  entitled  The  Elastic  Strength  of 
Guns,  may  be  consulted  for  detailed  discussions.  The  former 
gives  the  methods  and  formulas  for  navy  guns,  while  the  latter 
gives  those  for  army  guns;  these  differ  mainly  in  that  the  Navy 
employs  the  formulas  of  Clavarino,  while  the  Army  uses  those 
of  Birnie. 

Prob.  1550.  Prove  that  a  gun  tube  with  one  hoop  is  most  advan- 
tageously designed  when  the  common  radius  of  tube  and  hoop  is  a 
mean  proportional  between  the  other  two  radii. 

Prob.  1556.  Discuss  a  gun  with  two  hoops  where  ^1=3.50,  f^Q-iS, 
^  =  11.25,  ^4  =  12.25  inches  and  which  is  to  be  under  a  powder  pressure 
of  50  ooo  pounds  per  square  inch.  Find  a  set  of  shrinkages  so  that 
the  compressive  stress  at  the  bore  when  the  gun  is  at  rest  shall  be  45  ooo 
pounds  per  square  inch;  and  so  that  the  tensile  stresses  at  the  bore 
and  at  the  inside  of  each  hoop  during  the  explosion  shall  be  40  ooc 
pounds  per  square  inch. 


ART.  156  CYLINDRICAL  ROLLERS  403 


CHAPTER  XVII 
ROLLERS,   PLATES,    SPHERES 

ART.  156.     CLYINDRICAL  ROLLERS 

Let  cylindrical  rollers  of  diameter  d  and  length  /  be  placed 
between  two  flat  plates  and  transfer  a  load  from  the  upper  to 
the  lower  plate.  Fig.  156a  shows  end  and  side  views  of  the  two 
plates  with  one  roller  which  carries  the  load  W.  It  was  found 
in  the  experiments  of  Bach  that  the  plates  were  but  little  de- 
formed in  comparison  with  the  roller,  and  hence  the  entire  defor- 


iw       i   i    i    i   i   i  \ 


w 


-jw      ^    r  t    t   t    t  \w 

Fig.  156a  Fig.  1566 

mation  will  here  be  regarded  as  confined  to  the  latter.  The 
vertical  diameter  A  A  is  shortened  to  BB  and  any  vertical  chord 
aa  is  shortened  to  bb.  The  change  of  length  in  the  vertical  radius 
is  AB  and  that  in  the  vertical  half -chord  is  ab.  The  unit-shortening 
of  the  vertical  radius  is  AB/^AA  and  that  of  the  vertical  half- 
chord  is  ab/\aa  and  the  compressive  unit-stresses  are  propor- 
tional to  these  unit-shortenings  (Art.  10).  Let  the  greater 
shortening  AB  be  called  e  and  the  shortening  ab  be  called  y\ 
and  let  the  compressive  unit-stresses  at  B  and  b  be  called  S  and 
Sy.  Then,  if  the  elastic  limit  of  the  material  is  not  exceeded, 
S/SV  --=  e/y  or  Sy  =  S  .  y/e.  The  unit-stress  S  is  evidently  the 
maximum  and  it  is  required  to  determine  its  value  in  terms  of 
W,  d,  and  /. 

The  value  of  S  may  be  expressed  by  noting  that  e/\d  is  the 
unit-shortening  of  the  vertical  radius*  and  that  this  is  equal  to 


404  ROLLERS,  PLATES,  SPHERES  CHAP,  xvil 

S/E,  where  E  is  the  modulus  of  elasticity  (Art.  9) ;  hence  S/E 
=  c/\d.  The  sum  of  the  vertical  stresses  in  each  cylindrical 
segment  must  equal  the  total  load  W,  since  it  holds  that  load 
in  equilibrium.  Let  x  be  the  distance  Bb\  then  the  unit-stress 
Sy  acts  over  the  area  Idx,  and  hence  the  sum  of  all  the  vertical 
stresses  Sv  -  Idx  equals  W.  Accordingly, 

S/E  =  e/%d  and  t Sy  .  ldx=W 

are  two  equations  for  determining  the  values  of  S  and  e. 

To  solve  these  equations,  Sy  is  to-  be  replaced  by  its  value 
S.y/e   and   the   second  equation  then    becomes   Slj  ydx  =  We. 

Now  j ydx  is  the  area  of  the  circular  segment  CACBC\  but,  since 

the  deformation  is  very  slight,  the  arc  CAC  may  be  regarded  as 
parabolic  or  the  area  of  the  segment  as  $CCXAB.  Now  AB  =  t 
and  \CC=BC  =  (ed-e*)*=  (a*)*  nearly.  The  solution  of  the  twc 
equations  then  leads  to  the  formula, 


the  first  of  which  may  be  used  for  computing  the  unit-stress  5 
when  the  load  and  the  dimensions  of  the  roller  are  given,  while 
the  second  may  be  used  for  determining  the  size  of  a  roller  to 
carry  a  given  load  under  an  assigned  unit-stress. 

Let  w  be  the  load  per  unit  of  length  of  the  roller,  or  w  =  W/l, 
then  the  formula  (156)  may  be  written, 

W=$ldS(2S/E)*  or  w=idS(2S/E)*  (156)' 

which  shows  that  the  load  on  a  cylindrical  roller  should  vary 
directly  as  its  diameter.  Taking  5-15000  and  £  =  30000000 
pounds  per  square  inch  for  steel,  the  last  formula  reduces  to 
20  =  31 5(2.  This  agrees  well  with  the  rule  for  bridge  rollers  given 
in  Cooper's  Specifications  of  1901,  which  is  w  =  ^ood.  The  erro- 
neous rule,  w  =  i  2ooVd,  which  requires  the  load  to  vary  as  the 
square  root  of  the  diameter,  is  still  to  be  found  in  some  bridge 
specifications. 


ART.  156  CYLINDRICAL   ROLLERS  405 

As  a  numerical  example,  let  it  be  required  to  find  the  factor  of 
safety  of  six  wooden  rollers  used  in  moving  a  large  block  of  stone 
which  weighs  12  ooo  pounds,  the  diameter  of  each  roller  being 
6  inches  and  its  length  8  feet  4  inches.  Here  W  =  JXi2  000  = 
2000  pounds,  £  =  1500000  pounds  per  square  inch,  /  =  ioo 
inches,  and  d  =  6  inches;  then  formula  (156)  gives  5  =  266  pounds 
per  square  inch,  and  hence  the  factor  of  safety  is  8  000/266  =  30, 
which  indicates  a  high  degree  of  security.  Again,  let  it  be  re- 
quired to  find  the  diameter  of  a  cast-iron  roller  which  is  6  feet 
long  in  order  to  carry  30  ooo  pounds  with  a  factor  of  safety  of 
1  8.  Here  W  =  ^oooo  pounds,  5  =  90000/18  =  5000  pounds  per 
square  inch,  £  =  15  ooo  ooo  pounds  per  square  inch,  and  /  =  72 
inches;  then  in  the  formula  everything  is  known  except  d  and 
the  solution  gives  d  =  4.6  inches. 

The  assumption  that  the  plates  are  not  deformed  at  the  sur- 
face of  contact  with  the  roller  is  one  that  is  not  universally  ac- 
cepted. Later  experiments  by  Juselius  appear  to  indicate,  for 
rollers  and  plates  of  the  same  material,  that  the  deformation  of 
the  two  plates  in  any  vertical  is  about  equal  to  that  of  the  same 
vertical  in  the  roller.  By  using  this  conclusion,  as  indicated  in 
Fig.  1566,  the  above  reasoning  and  formulas  will  be  modified. 
The  shortening  of  the  vertical  radius  will  now  be  one-half  of  its 
former  value,  and  thus  the  first  formulas  of  (156)  and  (156)' 
become, 


Accordingly  the  compressive  unit-stress  due  to  a  given  load  is 
21  percent  less  than  before,  while  the  load  that  may  be  carried 
with  a  given  unit-stress  is  41  percent  greater  than  before.  Apply- 
ing this  second  formula  to  the  cast-iron  roller  of  the  last  para- 
graph, its  diameter  is  found  to  be  ^  =  3.4  inches.  It  is  seen, 
therefore,  that  the  assumption  used  at  the  beginning  of  this 
article  errs  on  the  side  of  safety  when  the  plates  are  actually 
deformed. 

Prob.  156.  A  load  of  192  ooo  pounds  is  carried  on  cylindrical  steel 
rollers  16  inches  long  and  3  inches  diameter.  Compute  the  number 
of  rollers  needed  when  the  allowable  unit-stress  is  12  ooo  pounds  per 
square  inch,  using  the  formula  which  appears  to  be  most  safe. 


406  ROLLERS,  PLATES,  SPHERES  CHAP.  XVTi 

ART.  157.     SPHERICAL  ROLLERS 

Let  a  sphere  of  diameter  d  be  placed  between  two  plates 
and  be  subject  to  compression  by  a  load  W.  The  left-hand 
diagram  of  Fig.  156a  may  represent  a  vertical  section  of  the  sphere 
and  plates,  the  former  being  regarded  as  alone  deformed.  The 
vertical  diameter  A  A  is  shortened  to  BB  and  any  vertical  line  aa 
is  shortened  to  bb.  Let  S  and  Sy  be  the  compressive  unit-stresses 
in  the  vertical  lines  A  A  and  aa;  let  the  greatest  shortening  AB 
be  called  e,  and  the  shortening  ab  be  called  y.  Then,  for  stresses 
within  the  elastic  limit,  S/Sy  =  e/y.  The  unit-stress  S  is  the 
greatest  and  it  is  required  to  find  its  value  in  terms  of  W  and  d. 

As  c/\d  is  the  unit-change  in  length  of  the  vertical  radius,  the 
value  of  S  may  be  expressed  by  S/E  =  e/^d  (Art.  10)  and  this 
is  one  equation  between  5  and  e.  To  find  another  equation, 
the  sum  of  the  veitical  stresses  in  the  spherical  segment  must 
equal  the  load  W.  Let  x  be  the  distance  Bb\  then  the  unit- 
stress  Sy  acts  over  the  area  2xx .  dx,  and  hence  the  sum  of  all 
the  vertical  stresses  Sy  .  znxdx  equals  W.  Accordingly, 

S/E  =  e/\d  2nfsv  .xdx=W 

are  two  equations  for  determining  the  values  of  S  and  e. 

To  solve  these  equations,  Sy  is  to  be  replaced  by  its  value 
S.y/e  and  the  second  equation  then  becomes  Sj2nyxdx=Wet 
Now  fznyxdx  is  the  volume  of  the  spherical  segment  whose 
section  is  CACBC  in  the  figure;  but,  since  the  deformation  is 
very  slight,  the  arc  CAC  may  be  regarded  as  parabolic,  and  then 
the  volume  is  one-half  that  of  a  cylinder  having  the  radius  BC 
and  the  altitude  AB.  Now  AB  =  e,  and  BC=(ed-e2)*  =  (ed)*, 
very  nearly,  since  e  is  small  compared  with  d.  Accordingly  the 
value  of  the  integral  is  %xe2d,  and  then  \nSed=W.  Inserting 
in  this  the  value  of  e  from  the  first  equation,  it  reduces  to, 

S2  =  WE/frd2  or  W=frd2S2/E  (157) 

From  the  first  formula  S  may  be  computed  when  W  is  given, 
and  from  the  second  W  may  be  computed  when  5  is  given.  The 
diameter  required  for  a  sphere  to  carry  a  given  load  with  an  allow- 


ART.  158    CONTACT  OF  CONCENTRATED  LOADS       407 

able  unit-stress  is  found  from  d2  =  ^WE/nS2 ;  thus  diameters  of 
spherical  rollers  should  vary  as  the  square  roots  of  their  loads. 

In  strictness  there  is  always  some  deformation  of  the  plates 
as  well  as  of  the  spheres,  and  the  old  assumption  that  the  total 
deformation  is  equally  divided  is  probably  nearer  the  truth  than 
that  it  is  all  confined  to  the  sphere.  Under  this  assumption  AB 
is  to  be  taken  as  \e  and  then  the  formulas  become, 

S2  =  WE/frd2,  W=&d2S2/E,  d2  =  2WE/nS2     (157)' 

Comparing  these  with  the  previous  formulas  it  is  seen  that  (157) 
give  values  of  S  which  are  41  percent  higher  than  (157)',  values 
of  W  which  are  only  one-half  as  large,  and  values  of  d  which  are 
41  percent  larger.  The  common  formulas  (157)  hence  err  on 
the  side  of  safety,  and  the  truth  probably  lies  between  them  and 
(157)'.  When  the  plates  are  harder  than  the  rollers,  (157)  is 
more  nearly  correct;  when  they  are  of  equal  hardness,  perhaps 
(157)'  gives  the  more  accurate  results. 

These  formulas,  like  those  of  the  last  articles,  are  valid  only 
when  the  load  produces  a  unit-stress  S  which  is  less  than  the 
elastic  limit  of  the  material.  For  stresses  beyond  the  elastic 
limit,  the  formulas  W  =  C\ld  for  cylinders  and  W^Ctffi  may 
be  considered  as  approximate,  in  which  C\  and  €2  are  to  be 
determined  by  experiment  for  each  material.  The  experiments 
of  Grand  all  and  Mars  ton  on  steel  cylindrical  rollers,  which  ranged 
in  diameter  from  i  inch  to  16  inches,  show  that  their  crushing 
loads  are  closely  given  by  the  formula  W  =  88old,  where  W  is  in 
pounds  and  /  and  d  in  inches. 

Prob.  157.  How  many  steel  spheres  are  required  to  carry  a  load 
of  6  ooo  pounds,  with  a  working  stress  of  1 5  ooo  pounds  per  square 
inch,  when  they  are  4  inches  in  diameter?  How  many  are  required 
when  they  are  12  inches  in  diameter? 

ART.  158.     CONTACT  OF  CONCENTRATED  LOADS 

When  a  concentrated  load  is  placed  upon  a  horizontal  beam  or 
plate,  it  produces  compressive  stresses  over  a  certain  area.  In 
bridges  and  buildings  concentrated  loads  are  often  applied  to 


408  ROLLERS,  PLATES,  SPHERES  CHAP,  xvil 

the  upper  surface  of  a  beam  by  means  of  another  beam  at  right 
angles  to  it;  in  this  case  the  surface  of  contact  is  plane  and  the 
concentrated  load  W  may  be  regarded  as  uniformly  distributed 
over  the  area.  This  subject  has  already  been  mentioned  in  Art. 
142,  and  it  is  there  indicated  that,  for  simple  beams,  the  flexural 
compressive  stress  and  the  direct  compressive  stress  due  to  the 
concentrated  load  combine  to  produce  a  true  compressive  stress 
which  is  smaller  than  either  of  them.  When  the  concentrated 
load  rests  upon  the  lower  flange  of  a  simple  I  beam,  as  sometimes 
occurs  in  practice,  the  combination  of  the  direct  compression 
with  the  flexural  tension  produces  a  true  compression  and  a  true 
tension  which  are  larger  than  the  apparent  ones.  It  is  hence 
always  preferable  to  support  the  concentrated  load  on  the  com- 
pressive side  of  a  beam. 

The  two  preceding  articles  contain  examples  of  the  contact 
of  cylinders  and  spheres  with  plane  surfaces,  and  from  the 
reasoning  there  given  a  relation  may  be  deduced  between  the 
area  of  contact  and  the  load  W.  For  the  cylinder  the  area  of 
contact  is  the  width  2(ed)*  multiplied  by  the  length  /;  inserting 
the  value  of  e,  this  area  is  a=2ld(S/2E)*,  and  replacing  S  by 
its  value  in  terms  of  W,  it  becomes  a=(ld)*(3W/E)*.  The 
area  of  contact  hence  varies  as  the  cube  root  of  the  load  for  the 
same  cylindrical  roller;  thus  if  a  load  W\  gives  an  area  of  contact 
fli,  a  load  &W'i  is  required  in  order  to  make  the  area  2d\.  This 
conclusion  is  valid  only  when  the  elastic  limit  of  the  material  is 
not  exceeded.  The  formula  here  deduced  for  a  is  for  the  case 
where  the  plate  is  not  deformed ;  when  the  deformation  is  equally 
divided  between  the  plate  and  the  roller,  $W  is  to  be  replaced 
by  6Wj  and  the  law  connecting  a  and  W  remains  unaltered. 

For  the  case  of  a  sphere  resting  on  a  plane,  Art.  157  shows 
that  the  area  of  contact  is  ned;  placing  in  this  the  value  of  e  in 
terms  of  5,  and  then  that  of  S  in  terms  of  W,  there  is  found 
a  =  d(nW/E)l,  which  shows  that  the  area  of  contact  varies  as 
the  diameter  of  the  sphere  and  with  the  square  root  of  the  load. 
To  double  the  area  of  contact,  it  is  hence  necessary  to  quad- 
ruple the  load  upon  a  sphere;  this  law  holds  whether  the  sphere 


AJIT.  159       CIRCULAR  PLATES  WITH  UNIFORM  LOAD  409 

alone  be  deformed  or  whether  the  deformation  is  divided  between 
the  sphere  and  plate.  The  above  law  does  not  agree  with  the 
conclusions  derived  from  the  experiments  made  by  J.  B.  Johnson 
on  the  contact  between  car- wheels  and  railroad  rails  (Art.  142). 
This  disagreement  is  probably  mostly  due  to  the  fact  that  the 
upper  surface  of  the  rail  is  not  a  plane,  and  in  part  to  the  fact 
that  the  unit-stresses  were  very  high. 

Prob.  158.  Compute  the  total  area  of  contact  for  the  cylindrical 
rollers  of  Problem  156.  If  the  same  load  is  carried  on  spherical  steel 
rollers  3  inches  in  diameter,  compute  the  total  area  of  contact. 

ART.  159.     CIRCULAR  PLATES  WITH  UNIFORM  LOAD 

Let  a  circular  plate  of  radius  r  and  uniform  thickness  d  be 
subject  on  one  side  to  a  pressure  R  on  each  square  unit  of  area, 
and  be  supported  or  fixed  around  the  circumference.  The  head 
of  a  cylinder  undei  the  pressure  of  water  or  steam  is  a  circular 
plate  in  such  a  ccndition.  Under  the  action  of  the  load,  the 
plate  bends,  the  side  in  contact  with  the  load  being  subject  to 
compression  while  the  other  side  is  under  tension;  the  maximum 
stress  caused  by  the  flexure  will  evidently  occur  at  the  middle, 
q,nd  this  is  required  to  be  determined. 

As  the  simplest  case  let  the  plate  be  merely  supported  around 
the  circumference.  The  total  load  on  the  plate  being  xr2R, 
the  total  reaction  of  the  support  is  also 
xr2R,  or  the  reaction  per  linear  unit  is  %rR. 
Now  let  a  strip  having  the  small  width  b 
be  imagined  to  be  cut  out  of  the  plate,  so 
that  its  central  line  coincides  with  a  diame- 
ter. The  reaction  at  each  end  of  this  strip 
is  b  .  \rR  and  the  load  on  the  strip  is  b  .  2rR. 
The  sum  of  the  two  reactions  being  only 
one- half  the  load,  an  upward  shearing  force 
equal  to  brR  must  act  along  the  sides  of  the 

strip  to  maintain  the  equilibrium.  The  manner  of  distribution 
of  this  shearing  stress  along  the  sides  of  the  strip  is  unknown 
and  uncertain,  but  a  fair  probable  assumption  may  be  to  take 


410  ROLLERS,  PLATES;  SPHERES  CHAP.  XVII 

it  as  constant  from  the  center  to  the  circumference  so  that  it  acts 
like  an  upward  uniform  load. 

The  strip  of  breadth  b,  depth  dt  and  length  2r  is  thus  a  simple 
beam  acted  upon  by  two  vertical  reactions,  each  equal  to  %brR,  a 
downward  uniform  load  2brR,  and  two  vertical  shears  on  the 
sides,  each  equal  to  \brR.  The  bending  moment  at  the  middle 
of  this  imaginary  beam  hence  is, 


and  the  maximum  unit-stress  on  the  upper  or  lower  fiber  at 
the  middle  of  the  strip  is,  from  the  flexure  formula  (41), 

S  =  Me  1  1  =  6M/bd2  =  f  £  .  r2/d2 

This  value  of  5  is  not  the  real  horizontal  unit-stress  at  the  center 
of  the  circle,  but  only  the  apparent  stress  due  to  considering  the 
elementary  strip.  At  the  center  the  horizontal  unit-stresses  are 
acting  in  all  directions.  If  a  second  strip  is  passed  in  Fig.  159 
at  right  angles  to  the  first,  a  unit-stress  S  equal  in  value  but  normal 
in  direction  to  the  first  will  be  found.  The  true  horizontal  unit- 
stress  T  will  be  determined  from  the  principle  of  Art.  139,  taking 
into  account  the  factor  of  lateral  contraction  X,  on  the  upper 
side  of  the  plate  T  =  S-kS-XR,  and  on  the  lower  side  of  the 
plate  r  =  5-A5.  The  latter  value  is  the  one  to  be  used,  since 
it  is  larger  than  the  former.  Accordingly, 

r=|(i  -X)R  .  (r/d)2  and  (d/r)2°=f  (i  -X)R/T 

are  the  general  formulas  for  the  discussion  of  circular  plates 
supported  around  the  circumference  and  subject  to  a  uniform 
load  nr2R. 

For  cast  iron  the  mean  value  of  the  factor  of  lateral  contrac- 
tion A  is  J,  while  for  wrought  iron  and  steel  it  is  J.     Hence, 

T=l(r/d)2R  and  T=(r/d)2R  (159) 

are  the  practical  formulas  for  use,  the  first  applying  to  cast  iron 
and  the  second  to  wrought  iron  and  steel  circular  plates  when 
supported  at  the  circumference,  T  being  the  allowable  unit-stress 
in  tension.  The  unit-pressure  R  that  a  circular  plate  can  carry 
varies  directly  as  the  square  of  its  thickness  and  inversely  as  the 
square  of  its  diameter. 


ART.  159          CIRCULAR  PLATES  WITH   UNIFORM  LOAD  411 

Another  method  of  discussing  this  case  is  to  consider  the  plate 
to  be  cut  along  a  diameter;   the  load  on  the  semicircle  is 
and  this  may  be  considered  as  acting 
at  its  center  of   gravity  clt   which  is 
distant  4^/3^  from  the  diameter  AC\ 
the  reaction   around  the  semicircum- 
ference  ABC  is  also  ^irr2R}  which  may 
be  considered  as  acting  at  its  center  of  Fig.  159 

gravity  c2  which  is  distant  2r/ir  from 

the  diameter  AC.  The  bending  moment  with  respect  to  this 
diameter  then  is 

M=  ±Trr*RX2r/Tr-  ^RX^r/^ir  =  %r*R. 

This  bending  moment  produces  flexural  stresses  in  the  section 
area  along  the  diameter  AC,  and  the  average  unit-stress  is  S'  = 
6M/b'd2,  where  b'  =  2r  and  d  is  the  depth  of  the  plate.  Repre- 
senting the  stresses  along  AC  by  the  ordinates  of  a  parabola,  the 
maximum  unit-stress  at  the  center  is  f  of  the  average,  or 


which  agrees  with  that  found  i»  the  previous  method.    Then, 
as  before,  the  true  unit-stress  is 


and  the  special  formulas  (159)  immediately  follow. 

The  more  common  case  of  a  circular  plate  fixed  around  its 
circumference  cannot  be  solved  without  a  discussion  of  the  elastic 
curve  into  which  a  diameter  deflects.  The  investigation  is  too 
lengthy  to  be  given  here,  but  it  can  be  said  that  the  true  effective 
unit-stress  is  about  two-thirds  of  that  for  the  supported  plate. 

Hence 

T=l(r/d)2R  and  T=$(r/d)*R  (159)' 

are  formulas  for  fixed  plates,  the  first  being  for  cast  iron  and  the 
second  for  wrought  iron  and  steel. 

From  the  above  formulas  the  proper  thickness  d  for  circular 
plates  under  uniform  pressure  may  be  readily  computed.  For 
example,  let  a  fixed  cast-iron  cylinder  head  of  36  inches  diameter 
be  required  to  carry  a  uniform  pressure  R  of  250  pounds  per 
square  inch,  with  an  allowable  tensile  stress  T  of  3  600  pounds 


412  ROLLERS,  PLATES,  SPHERES  CHAP.  XVII 


per  square  inch;  then  d  =  r(^R/4.T)^  =  4.1  inches.  The  thickness 
of  a  steel  cylinder  head  for  the  same  diameter  and  pressure,  for 
a  tensile  stress  of  12  ooo  pounds  per  square  inch,  will  be 
2*i  inches. 


Prob.  159a.  When  the  total  load  W  for  a  circular  plate  is  given, 
show  that  the  thickness  of  the  plate  should  be  the  same  whatever  be  the 
diameter. 

Prob.  1596.  If  a  plate  36  inches  in  diameter  and  2  inches  thick 
can  safely  carry  a  pressure  of  250  pounds  per  square  inch,  what  is  the 
safe  pressure  for  a  plate  24  inches  in  diameter  and  i  inch  thick  ? 

ART.  160.     CIRCULAR  PLATES  UNDER  CONCENTRATED  LOAD 
When  a  circular  plate  is  under  flexure  from  a  concentrated 
load  at  the  middle,  it  is  more  highly  stressed  than  when  the  same 

load  is  uniformly  distributed.  The 
following  approximate  discussion  of 
this  case  is  for  a  circular  plate  sup- 
ported along  its  circumference,  where 
the  load  P  is  uniformly  distributed  over 
Fig.  160  a  circle  of  radius  ro,  the  radius  of  the 

plate  being  r  and  its  thickness  d.    Let 

Pig.  160  represent  one-half  of  the  plate,  DEF  being  the  semicircle 
whose  radius  is  TQ.  The  load  ^P  being  uniformly  distributed,  the 
distance  of  its  center  of  gravity  from  the  diameter  A  C  is  47-0/371-; 
the  reaction  \P  along  the  circumference  ABC  has  its  center  of 
gravity  at  the  distance  2r/ir  from  AC.  The  bending  moment  with 
respect  to  the  diameter  AC  then  is 

—  ~—  - 

7T  37T 

This  bending  moment  produces  flexural  stresses  in  the  section 
area  along  the  diameter  AC,  and  the  average  unit-stress  is  Sr  = 
6M/b'd2  where  bf  =  2r.  If  the  stresses  normal  to  AC  are  rep- 
resented by  the  ordinates  of  a  parabola,  the  maximum  unit-stress 
at  the  center  is  f  of  the  average,  or 


ART.  160    CIRCULAR  PLATES  UNDER  CONCENTRATED  LOAD      413 

If  a  second  diameter  be  considered  normal  to  AC,  a  unit-stress 
S  equal  to  this  is  also  found,  and  from  Art.  139,  taking  into 
account  the  factor  of  lateral  contraction  X,  the  true  unit-stress 
on  the  lower  side  of  the  plate  is  T=  (i  -X)S,  or 

(160) 

in  which  A  is  J  for  cast  iron,  J  for  wrought  iron  and  steel,  and 
about  I  for  concrete. 

When  the  load  covers  the  entire  plate,  then  ro  =  r  and  (160) 
becomes  r  =  f(i  -tyP/ird2;  in  this  .put  P  =  irr2R,  where  R  is 
the  uniform  unit-pressure  due  to  P,  and  it  reduces  to  the  same 
formula  as  derived  in  Art.  159.  When  P  is  concentrated  at  the 
middle  of  the  plate,  then  r0  =  o  and  (160)  becomes  r=f  (i  -X)P/ird2. 
Hence  a  load  concentrated  at  the  middle  of  a  circular  plate  pro- 
duces three  times  the  stress  of  that  due  to  the  same  load  uniformly 
distributed.  As  a  matter  of  fact  it  would  be  impossible  to  con- 
centrate P  at  a  mathematical  point,  since  then  the  unit-pressure 
R  which  is  P/irro2,  would  be  infinite;  in  cases  of  design  R  should 
not  exceed  the  elastic  limit  of  the  material. 

As  an  example  let  a  load  of  6  ooo  pounds  be  at  the  middle  of  a 
steel  plate,  distributed  over  a  circle  i  inch  in  diameter,  the  plate 
being  finch  thick  and  24  inches  in  diameter;  here  A  =  J,  ro  =  i 
inch,  r=  12  inches,  and  d  =  }  inch.  Then  formula  (160)  gives  T= 
9  950  pounds  per  square  inch  as  the  true  tensile  unit-stress  at  the 
middle  of  the  lower  side  of  the  plate,  which  is  a  safe  allowable  value. 

When  a  plate  is  fixed  around  the  circumference  it  is  probable 
that  the  constant  in  (160)  should  be  3  instead  of  f .  Fixing  the 
circumference  increases  the  strength  of  the  plate  for  the  same 
reason  that  the  strength  of  a  beam  is  increased  by  fixing  its  ends. 
A  fixed  plate  can  carry  a  load  about  50  per  cent  greater  than  that 
carried  by  a  supported  one.  When  a  plate  is  stiffened  by  ribs, 
as  is  often  the  case  in  cast  iron,  about  one-half  of  the  material 
of  the  ribs  may  be  regarded  as  adding  to  the  thickness  of  the  plate. 

Prob.  160a.  Which  is  the  stronger,  a  circular  plate  carrying  a  load  P 
uniformly  distributed,  or  one  carrying  a  load  \P  which  distributed  over 
an  area  at  the  middle  which  is  one-third  of  the  area  of  the  plate? 


414  ROLLERS,  PLATES,  SPHERES  CHAP,  xvil 

ART.  161.     ELLIPTICAL  PLATES 

Elliptical  plates  are  commonly  used  for  the  covers  of  man- 
holes in  boilers  and  stand-pipes.  Let  R  be  the  uniform  unit- 
pressure  on  the  plate,  a  the  semi-major  axis,  and  b  the  semi-minor 
axis  of  the  ellipse.  It  is  required  to  find 
the  maximum  unit-stress  T  on  the  tensile 
side  of  the  plate. 

Taking  the  case  where  the  plate  is 
simply  supported  around  the  circumfer- 
ence, let  two  elementary  strips  be  drawn 

as  in  Fig.  161,  one  along  the  major  axis  and  the  other  along  the 
minor  axis.  Let  W\  and  Wz  be  the  loads  on  these  strips,  and 
fi  and  /2  their  deflections.  At  the  center  of  the  ellipse  the  deflec- 
tions of  the  strips  are,  from  Art.  55, 


and  because  these  are  equal,  TFi#3  must  be  equal  to  WJP.  Since 
the  reactions  at  the  ends  are  proportional  to  the  loads,  it  follows 
that  the  reactions  at  the  ends  of  the  axes  are  inversely  as  the 
cubes  of  the  lengths  of  the  axes.  Hence  the  total  weight  xabR 
is  not  uniformly  distributed  on  the  support  around  the  circum- 
ference, but  the  greatest  reaction  per  linear  unit  will  be  found 
at  the  ends  of  the  minor  axis  and  the  least  at  the  ends  of  the 
major  axis.  It  should  hence  be  expected  that  the  horizontal 
flexural  stresses  at  the  center  are  the  greatest  in  directions  parallel 
to  the  minor  axis,  and  that  in  case  of  rupture  a  crack  would 
begin  at  the  center  and  run  along  the  major  axis  ;  this  is  verified 
by  tests. 

The  theoretic  solution  of  this  very  difficult  problem  cannot 
well  be  given  here.  From  the  discussion  of  Grashof  and  the 
experiments  of  Bach  the  following  approximate  formula  may  be 
written  for  wrought-iron  and  steel  elliptical  plates  supported 
around  the  circumfeience: 


(161) 

in  which  a  and  b  are  the  semi-axes,  d  the  thickness  of  the  plate, 
R  the  unit-pressure  upon  it,  and  T  the  allowable  tensile  unit- 


AKTT.  162  RECTANGULAR    PLATES  415 

stress.     When  a  and  b  are  equal,  the  ellipse  becomes  a  circle  of 
radius  r,  and  T  =  R(r/d)2  as  found  in  Art.  159. 

For  a  cast-iron  plate  supported  along  its  circumference,  the 
numerical  coefficient  in  the  above  formula  will  be  f  instead  of  2. 
For  plates  fixed  around  the  circumference,  the  coefficient  will 
be  about  f  for  wrought  iron  and  steel,  and  about  ij  for  cast  iron. 
These  numbers  are  derived  by  taking  the  factor  of  lateral  contrac- 
tion as  J  for  wrought  iron  and  steel,  and  as  J  for  cast  iron.  In 
Germany  the  value  A  =  -&  is  generally  used  for  both  cast  iron 
and  steel,  and  this  will  slightly  modify  the  above  numerical 
coefficients. 

A  common  proportion  for  manhole  covers  is  to  make  0/6  =  1.5,, 
that  is,  the  length  is  50  percent  greater  than  the  width.  Let 
the  length  be  24  inches  and  the  width  16  inches,  and  let  it  be 
required  to  find  the  proper  thickness  when  the  cast-iron  manhole 
cover  is  used  in  a  stand-pipe  under  a  head  of  water  of  50  feet. 
Here  a  =  12  and  6=8  inches,  R  =  22  pounds  per  square  inch, 
while  the  allowable  T  for  cast  iron  may  be  taken  as  3  ooo  pounds 
per  square  inch.  .Then,  since  the  plate  is  supported  around 
the  circumference,  the  numerical  coefficient  in  the  formula  will 
be  f,  and  from  it  is  found  d=ab(%R/(a2  +  b2)T)*  =  o.86  inches, 
so  that  a  thickness  of  J  inches  will  be  sufficient  safely  to  with- 
stand the  pressure. 

Prob.  1610.  Show  that  the  allowable  unit-pressure  R  for  a  cast- 
iron  elliptical  manhole  cover  having  the  proportions  0/6=1.5,  1S 
given  by  R  =  2.g2(d/bi)2T,  in  which  b\  is  the  width  of  the  plate. 

Prob.  1616.  Compute  the  safe  unit-pressure  for  a  cast  iron  man- 
hole cover  of  20  inches  length,  13  inches  width,  and  ij  inches  thick- 
ness. What  head  of  water  will  produce  this  pressure? 

ART.  162.     RECTANGULAR  PLATES 

A  rectangular  plate  of  length  2/  and  width  2m,  subject  to  a 
uniform  pressure  R  per  square  unit,  distributes  that  pressure 
over  the  support  in  a  similar  manner  to  the  elliptical  plate.  The 
reaction  per  linear  unit  is  less  on  the  ends  than  on  the  sides, 


416  ROLLERS,  PLATES,  SPHERES  CHAP.  XVII 

and  is  greater  at  the  middle  of  the  ends  and  sides  than  near  the 
corners.  Rupture  tends  to  occur  near  the  center  and  parallel 
to  the  longer  side.  The  approximate  formula  derived  from  the 
discussion  of  Bach  for  iron  and  steel  plates  is, 


2)d2  (162) 

in  which  T  is  the  maximum  tensile  unit-stress  at  the  middle  of  the 
lower  side  of  the  plate,  and  d  is  the  thickness  of  the  plate.  The 
values  of  the  number  a,  as  determined  by  the  experiments  of 
Bach,  ranged  from  J  to  f  ,  according  as  the  condition  of  the  edges 
approached  that  of  a  mere  support  or  a  state  of  fixedness. 

For  a  square  plate  /  and  m  are  equal,  and  the  above  expression 
may  then  be  reduced  to  the  forms, 

T=%R(l/d)2  and  T=$R(l/d)2 

the  first  being  for  free  and  the  second  for  fixed  edges.  The 
numerical  constants  in  these  formulas  are  derived  from  the  dis- 
cussion of  experiments  and  hence  stand  upon  a  different  basis 
from  those  deduced  for  circular  plates;  probably  the  formulas 
will  apply  better  to  cases  of  rupture  than  to  cases  where  T  is 
within  the  elastic  limit  of  the  material. 

This  problem  has  been  discussed  theoretically  by  Grashof  with 
the  conclusion  that  the  formula  for  a  square  plate  fixed  at  the 
middle  of  each  edge  is  T  =  (i-tf)R(l/d)2,  where  X  is  the  factor 
of  lateral  contraction.  A  plate  might  be  fixed  in  this  manner 
by  a  bolt  at  the  middle  of  each  edge,  but  such  an  arrangement 
is  unusual,  the  common  method  being  to  bolt  it  to  the  support 
at  many  points.  When  A  =  J,  this  formula  reduces  to  T  =f  ^(//d)2, 
which  is  intermediate  between  those  given  for  free  and  for  fixed 
edges  in  the  last  paragraph. 

While  the  numerical  coefficients  for  square  plates,  as  deduced 
by  different  authors,  vary  somewhat,  it  is  well  established  that 
the  unit-stress  T  at  the  middle  of  the  plate  varies  directly  as  its 
area  and  the  unit-pressure  R,  and  inversely  as  the  square  of  its 
thickness.  The  strength  of  a  square  plate,  as  measured  by  the 
pressure  R  that  it  can  carry,  varies  directly  as  the  square  of  the 


ART.  163  HOLLOW  SPHERES  417 

thickness  and  inversely  as  the  area;  this  law  is  the  same  as  that 
previously  found  for  circular  plates. 

Prob.  162.  Prove  that  the  maximum  unit-stress  for  a  square  plate, 
caused  by  a  given  uniform  load  W,  is  independent  of  the  size  of  the 
plate. 

ART.  163.     HOLLOW  SPHERES 

Hollow  spheres  are  used  in  certain  forms  of  boilers  under 
inside  steam-pressure.  The  ends  of  steam  and  water  cylinders 
are  sometimes  made  hemispherical  instead  of  plane,  in  order 
to  avoid  flexure;  the  base  of  a  steel  water  tank  is  often  made  a 
hemisphere  for  the  same  reason.  If  the  thickness  of  the  sphere 
is  small  compared  to  its  radius,  the  investigation  is  simple.  Let 
r  be  the  radius  and  /  the  thickness.  Let  R  be  the  inside  pressure 
per  square  unit,  and  S  the  tensile  unit-stress  on  the  annulus. 
Then  on  any  great  circle  the  total  pressure  is  xr2R,  and  this  is 
resisted  by  the  tension  inrtS  in  the  section  of  the  annulus.  By 
equating  these,  there  is  found, 

2tS  =  rR  or  S=%R.r/t 

which  is  the  formula  generally  used  for  thin  spheres  under  inner 
pressure.  But  in  strictness  S  is  the  apparent  stress,  while  another 
equal  in  intensity  acts  at  right  angles  to  it.  Thus  from  Art.  139 
the  true  stress  on  the  outside  surface  isTi=S-AS,  while  that 
on  the  inside  surface  is  7^2= 5  —  A5  +  A7?.  Using  J  for  the  value  of 
A,  and  inserting  the  above  value  of  S,  there  result, 

Ti  =  $R.r/t  and  T2  =  IR+$R .  r/t 

for  the  true  tensile  unit-stresses  on  the  outside  and  inside  surfaces 
respectively.  Both  of  these  are  less  than  5,  and  hence  the  usual 
formula  for  thin  spheres  (Art.  31)  errs  on  the  side  of  safety. 

The  investigation  of  a  thick  hollow  sphere  under  inside  and 
outside  pressure  will  be  similar  to  that  of  the  thick  cylinder  in 
Art.  149.  Let  r\  be  the  inside  and  r^  the  outside  radius,  R\  and 
i?2  being  the  corresponding  pressures  per  square  unit  of  surface. 
Fig.  1496  may  represent  a  partial  section  of  the  sphere,  x  being 
the  radius  of  any  elementary  annulus  where  the  radial  unit-stress 


418  ROLLERS,  PLATES,  SPHERES  CHAP.  XVII 

is  R  and  the  tangential  unit-stress  is  S.  From  the  symmetry 
of  the  sphere  it  is  seen  that  another  stress  S  acts  at  right  angles 
to  the  one  shown  in  the  figure.  Thus  an  elementary  particle 
at  any  position  in  the  annulus  is  held  in  equilibrium  by  three 
principal  stresses  R,  S,  and  S.  The  sum  of  these  is  regarded 
as  constant  throughout  the  annulus  (Art.  183),  and  accordingly 
2S  +  R=sCi  is  one  equation  between  S  and  R,  where  C\  is  a 
constant  which  is  to  be  determined. 

Now  the  inside  pressure  on  a  great  circle  whose  radius  is  x 
is  nx2R,  and  the  outside  pressure  on  a  great  circle  whose  radius 
is  x  +  dx  is  x(x  +  dx)2(R  +  dR),  both  of  these  being  perpendicular 
to  the  plane  of  the  circle.  The  difference  of  these  is  equal  to 
the  resisting  stress  in  the  elementary  annulus,  which  is  2nxdx  .  S. 
Stating  this  equation  and  omitting  quantities  of  the  second  order, 
a  second  relation  between  5  and  R  is  found.  Accordingly, 


are  the  two  conditions  for  determining  S  and  R.  Substituting 
in  the  second  equation  the  value  of  S  from  the  first  and  inte- 
grating, the  value  of  R  in  terms  of  x  is  found,  and  then  that  of 
R  is  known;  thus, 

S  =  Cl  +  C2/x*  and  R  =  d-2C2/x3  (163) 

in  which  C2  is  a  constant  of  the  integration.  These  formulas 
for  hollow  spheres  are  seen  to  be  analogous  to  those  for  thick 
cylinders,  the  radii  being  cubed  instead  of  squared.  The  formula 
for  S  is  the  most  important  one  and  S  has  its  greatest  value  at 
the  inside  surface  of  the  hollow  sphere. 

Values  of  the  constants  C\  and  C2  may  be  found  from  the 
formula  for  R.  Regarding  the  unit-pressures  RI  and  R2  as 
without  sign,  R  becomes  —Ri  when  x=r\  and  R  becomes  —  R% 
when  x=r2',  then, 

-  r<?R2)  /(rf  -  n3)  C2  =  Jn3r23(*i  -R2)/(r2*  -  n8) 


and  these  when  placed  in  (163)  give  the  formulas  deduced  by 
Lame  for  thick  hollow  spheres.  The  most  common  case  is  that 
where  there  is  no  outside  pressure  R2',  here  the  tensile  unit- 


ART.  163  HOLLOW  SPHERES  419 

stresses  on  the  inside  and  outside  surfaces  are, 

Si=£i(W+'i3)/(r23-ri3)  S2=*i  .|fi3/W-ri3)     (163)' 

and  the  first  of  these  gives  the  greater  unit-stress.  For  example, 
if  r2  =  2ri,  then  Si=^Ri,  and  when  r2  is  nearly  as  large  as  r\, 
then  Si  is  nearly  %R  .  r/t,  as  previously  found  for  a  thin  hollow 
sphere  in  Art.  31. 

The  above  gives  the  apparent  unit-stresses.  To  find  the 
true  unit-stress  T\  at  the  inside  surface  of  a  steel  hollow  sphere, 
the  factor  of  lateral  contraction  is  to  be  taken  as  J,  and  then  by 
Art.  139, 

ri=5i~J5i+t/?i=t^1(2f23+fi3)/(f23-ri8)  (163)" 

which  will  generally  be  found  to  be  less  than  Si.  It  is  therefore 
on  the  safe  side  to  use  the  formula  for  Si  in  cases  of  design. 

As  an  example,  let  a  steel  cylinder  4  inches  in  inside  and 
8  inches  in  outside  radius  have  a  hemispherical  end  with  the  same 
radii,  and  be  subject  to  an  inside  water  pressure  of  4  ooo  pounds 
per  square  inch.  Then  the  apparent  tensile  stress  on  the  inside 
surface  of  the  hemisphere  is  found  from  (163)'  to  be  2  860  pounds 
per  square  inch,  while  (163)"  gives  the  true  tensile  stress  as 
3  240  pounds  per  square  inch;  hence  the  true  stress  is  about 
13  percent  larger  than  the  apparent.  For  the  cylinder  itself, 
the  apparent  and  true  tensile  stresses  at  the  inside  surface  may 
be  computed  from  (150)  and  (152)"  and  these  values  are  Si  =6  700 
and  T"i=7  600  pounds  per  square  inch,  so  that  the  true  stress 
is  13  percent  greater  than  the  apparent  for  the  cylinder. 

If  the  end  of  this  cylinder  is  a  flat  plate  of  the  same  thickness 
as  the  cylinder,  or  4  inches,  and  fixed  around  the  circumference, 
the  true  stress  on  the  outer  side  is  found  from  (159)'  to  be 
r  =  13X4000  =  16  ooo  pounds  per  square  inch;  this  is  five  times 
as  great  as  that  for  the  hemisphere,  and  more  than  double  the 
greatest  stress  in  the  cylinder.  The  advantage  of  hemispherical 
ends  in  reducing  the  stresses  is  thus  seen  to  be  very  great.  It  may 
be  remarked,  in  conclusion,  that  the  theory  of  internal  stress  in 
cylinders  and  spheres  is  not  perfect,  for  it  fails  to  give  the  same 


420  ROLLERS,  PLATES,  SPHERES  CHAP.  XVII 

results  for  the  common  surface  of  junction  of  a  cylinder  and 
hemisphere.  This  indicates  that  the  assumption  made  regarding 
the  constancy  of  2$+R  does  not  probably  hold  good  for  a  hemi- 
sphere attached  to  a  cylinder. 

Prob.  163#.  A  hollow  sphere  is  to  be  subject  to  a  steam-pressure  of 
600  pounds  per  square  inch,  its  inner  radius  being  8  inches.  Compute 
its  thickness,  so  that  the  greatest  tensile  stress  may  be  i  ooo  pounds 
per  square  inch. 

Prob.  1636.  Investigate  the  discrepancy  between  the  formulas  for 
hollow  cylinders  and  hollow  spheres  for  the  following  numerical  case. 
A  hollow  cylinder  with  hollow  hemispherical  ends,  the  inside  diameters 
being  8  inches  and  the  outside  diameters  12  inches,  is  subject  to  an 
inside  water  pressure  of  2  400  pounds  per  square  inch.  Compute,  by 
Art.  152,  and  by  this  article,  the  true  maximum  unit-stress  T  for  the 
common  plane  of  junction  of  cylinder  and  hemisphere. 


ART.  164  CENTRIFUGAL   TENSION  421 

CHAPTER  XVIII 
MISCELLANEOUS    DISCUSSIONS 

ART.  164.     CENTRIFUGAL  TENSION 

WHEN  the  center  of  gravity  of  a  body  of  weight  P  revolves 
around  an  axis  with  the  uniform  velocity  Vi  and  r  is  the  distance 
of  the  center  of  gravity  from  the  axis,  there  is  generated  a  stress 
in  the  cord  or  bar  that  connects  the  body  with  the  axis.  This 
centrifugal  force  Q  is  shown  in  works  on  theoretical  mechanics 
to  be  Q  =Pv2/gr,  where  g  is  the  acceleration  of  a  body  falling 
vertically  under  the  action  of  gravity  near  the  surface  of  the  earth. 
The  case  shown  in  Fig.  1640  is  that  of  a  bar  of  uniform  section 
area  and  length  /,  the  weight  P  being  attached  to  one  end  while 
it  revolves  around  an  axis  A  at  the  other  end.  It  is  required 
to  find  the  centrifugal  stress  in  the  bar  at  A  when  the  speed  of 
n  revolutions  per  second  is  maintained. 

N 

A 


19 

I 

u*  .  — 

7                    /_ 

L 

r_               I 

Fig.  164a 

Let  r  be  any  distance  from  the  axis;  the  velocity  at  this  dis- 
tance is  mm,  or  if  co  is  the  angular  velocity,  the  velocity  v  at 
the  distance  r  is  raj,  and  the  relation  between  n  and  to  is  given 
by  (t)  =  27cn.  Now  let  W  be  the  weight  of  the  bar,  J/  the  distance 
of  its  center  of  gravity  from  the  axis,  and  v^  the  velocity  of  that 
center  of  gravity  ;  then, 


gives  the  centrifugal  force  at  the  axis  ;   this  produces  a  tension  in 
the  bar  and  a  sidewise  compression  and  flexuie  on  the  axis. 

As  an  example,  let  a  bar  of  wrought  iron  2X2  inches  and 
6  feet  long  have  a  weight  of  400  pounds  with  its  center  of  gravity 


422  MISCELLANEOUS  DISCUSSIONS  CHAP,  xvill 

6J  feet  from  the  axis  of  revolution.  It  is  required  to  find  the 
number  of  revolutions  per  second  in  order  to  produce  rupture. 
Solving  the  last  equation  for  w,  and  placing  Q=  50000  pounds 
per  square  inch,  P=4oo  pounds,  W  =So  pounds,  £  =  32.16  feet 
per  second  per  second,  1=6  feet,  and  r=6J  feet,  there  is  found 
^=48.4  radians  per  second,  and  hence  the  speed  required  to 
cause  rupture  is  n  =48.4/27:  =  7.  8  revolutions  per  second. 

When  P=o  in  the  above  formula,  the  case  is  that  of  a  bar 
of  uniform  section  area  and  of  weight  W,  and  the  tensile  stress 
at  the  axis  is  Q  =  %Wlco2/g.  The  tensile  stress  in  such  a  bar 
is  o  at  the  free  end  and  it  increases  toward  the  axis,  where  it  has 
the  value  Q.  Let  a  be  the  area  of  the  cross-section,  w  the  weight 
of  a  unit  of  volume,  and  x  any  distance  from  the  axis.  Then  the 
weight  of  the  length  l  —  x  is  wa(l—x)  and  the  distance  of  its 
center  of  gravity  from  the  axis  is  J(/  +  #),  so  that  the  centrifugal 
force  of  this  part  of  the  bar  is, 


and  this  is  the  tensile  stress  at  the  distance  x  from  the  axis.  When 
x=l,  then  Q'  =o;  when  x  =  o,  then  Q'  =  wal2aj2/2g,  which  is  equal 
to  the  above  value  of  Q.  The  tensile  unit-stress  in  the  bar  at 
any  distance  x  from  the  axis  is  R=Q'/a.  This  may  be  written 
R  =  $(wa?/g)(l2  —  x2)  and  it  will  be  seen  to  be  closely  analogous 
with  the  expression  for  radial  unit-stress  in  a  revolving  fly-wheel 
or  millstone. 

Another  case  is  that  of  the  thin  circular  hoop  of  mean  radius 
r  and  thickness  /,  shown  in  the  first  diagram  of  Fig.  164&.     Let 

W  be  its  weight,  which  is  equal 
to  2xwbrt,  if  w  is  the  weight  of 
the  material  per  cubic  unit,  and 
b  is  the  width  of  the  hoop  per- 
pendicular to  the  plane  of  the 
drawing.  The  centrifugal  force 
acting  on  the  axis  is  here  zero 
because  the  center  of  gravity  of  the  hoop  coincides  with  the  axis. 
There  is  hence  no  pressure  on  the  axis,  but  the  centrifugal  force 
acts  radially  upon  the  hoop  and  produces  tension  in  it  in  the 


ART.  164  CENTRIFUGAL  TENSION  423 

same  manner  as  inside  water  pressure  does  in  a  thin  pipe.  The 
centrifugal  force  due  to  an  angular  velocity  aj  is  2xivbr2ta)2  j g 
for  the  entire  weight  and  hence  the  radial  unit-pressure  Rf  is 
found  by  dividing  this  by  the  area  2nbr,  or  R' =  (wco2  /  g)rt.  Now, 
referring  to  Art.  30,  it  is  seen  that  rRf  —  IS  is  the  relation  between 
the  outward  unit-pressure  R'  and  the  tangential  unit-stress  S. 
Hence  the  value  of  the  latter  is 

S '=  (waj2  /  g)r2  w  =  27tn 

in  which  n  is  the  number  of  revolutions  per  second.  For  ex- 
ample, let  it  be  required  to  find  the  tensile  stress  in  a  cast-iron 
hoop  4  inches  wide,  2  inches  thick,  and  62  inches  in  outside 
diameter  when  making  300  revolutions  per  minute.  Here 
7^  =  450/1728  pounds  per  cubic  inch,  ^  =  30  inches,  w  =  5  revolu- 
tions per  second,  and  #  =  32.16X12  inches  per  second  per  second; 
then  S  is  600  pounds  per  square  inch. 

For  a  hoop  of  considerable  thickness  let  r\  be  the  inside 
and  T2  the  outside  radius.  Let  x  be  the  inside  and  x  +  dx  be 
the  outside  radius  of  any  elementary  annulus,  and  R  and  R  +  dR 
the  radial  unit-stresses,  these  being  considered  tensile,  as  in 
Art.  149.  The  outward  forces  acting  upon  the  annulus  are  the 
radial  unit-stress  R  +  dR  and  the  centrifugal  unit-pressure  Rf, 
the  value  of  which  was  found  above  to  be  (waj?/g)xdxt  while 
the  inward  force  is  the  radial  unit-stress  R.  Accordingly,  just 
as  for  a  thin  water-pipe,  the  equation  of  equilibrium  between 
these  forces  and  the  tangential  unit-stress  S  is 

(R'  +  R  +  dR) (x  +  dx)  -Rx  =  Sdx 

which,  neglecting  quantities  of  the  second  order  and  representing 
(waP/g)  by  m,  reduces  to  the  differential  equation 

mxldx + Rdx  +  xdR  =  Sdx 

and  this  must  be  satisfied  by  the  expressions  for  R  and  5.  It  is 
also  to  be  noted  that  for  a  solid  wheel,  for  which  fi=o,  the 
values  of  R  and  5  must  be  equal  when  x  =  o.  Further,  for  a 
hollow  wheel  or  hoop  the  value  of  R  must  be  zero  when  x  =  r2 
and  also  when  x  =  ri,  since  there  are  no  radial  stresses  on  the  free 
circumferences.  The  expressions 


424  MISCELLANEOUS  DISCUSSIONS          CHAP,  xvm 

satisfy  these  two  conditions  and  also  the  differential  equation, 
provided  q  =  m/(^—p).  In  order  to  determine  p,  take  the  case 
of  a  solid  wheel  for  which  r\  =o  and  r2  =  r,  the  formulas  becoming 

R  =  m(r2_x2)/(3_p)  S=m(r2-px2)/(3-p) 

and  let  p  be  determined  from  the  condition  that  the  total  internal 
work  shall  be  a  minimum  (Art.  126).  This  investigation  gives 
J  for  the  value  of  p,  whence  q  =  f  w,  and  thus  finally, 


are  the  radial  and  tangential  unit-stresses  at  the  distance  x  from 
the  center  of  a  revolving  wheel  of  uniform  thickness. 

These  formulas  show  for  a  solid  grindstone  or  millstone,  for 
which  ri=o  and  rz  =  r,  that  the  greatest  stress  occurs  at  the 
center,  R  and  S  being  each  equal  to  (wa)2  /  g)fr2  ,  while  for  the 
circumference  R  =  o  and  S  =  (wa)2  1  g)\r2  .  They  also  show,  for 
a  grindstone  or  millstone  having  a  hole  of  radius  r  \  at  the  center, 
that  the  greatest  value  of  R  occurs  for  x=(r\r^t  while  the 
greatest  value  of  S  occurs  at  the  inner  circumference.  It  is  hence 
to  be  expected  that,  under  a  sufficiently  high  velocity,  a  solid 
wheel  would  begin  to  crack  at  the  center  and  a  hollow  wheel  at 
the  inner  circumference. 

The  above  refers  to  apparent  stresses  only,  but  the  true 
stresses  corresponding  to  the  actual  deformations  are  somewhat 
less  than  given  by  the  formulas,  since  R  and  5  are  both  tensile. 
Let  X  be  the  given  factor  of  lateral  contraction  the  value  of  which 
is  J  for  steel,  J  for  cast  iron,  and  J  or  less  for  stone;  then  the  true 
radial  unit-stress  is  R  —  AS  and  the  true  tangential  unit-stress  is 
S—AR  in  which  R  and  S  are  given  by  the  preceding  formulas. 
It  should  be  noted,  however,  that  more  exhaustive  investigations 
seem  to  lead  to  somewhat  different  expressions.  Love's  formula 
for  the  tangential  true  unit-stress  is 


For  A  =  l  this  gives  S=  (waP/gjffir2  at  the  center  of  a  solid  wheel, 
while  the  preceding  formula  gives  S=(wa}2/g)%%r2,  which  is  8 
percent  greater. 


ART.  165  CENTRIFUGAL    FLEXURE  425 

ART.  165.     CENTRIFUGAL  FLEXURE 

The  rod  that  connects  the  cross-head  of  a  steam  engine  with 
the  crank  pin  is  subject  to  a  centrifugal  flexural  stress  owing 
to  the  fact  that  one  end  revolves  in  a  circle.  The  horizontal 
rod,  or  parallel  bar,  joining  two  driving  wheels  of  a  locomotive 
is  another  instance  of  centrifugal  flexure;  this  is  simpler  than 
the  connecting  rod,  because  all  points  are  revolving  with  the 
same  velocity,  and  hence  it  will  be  discussed  first. 

Let  u  be  the  velocity  of  a  locomotive  and  v  the  velocity  of 
revolution  of  the  end  of  the  parallel  rod  around  che  axle  of  the 
driving  wheel  to  which  it  is  attached.  Let  r\  be  the  radius  of 
the  wheel,  and  r  the  radius  of  the  circle  of  revolution  of  the  end 
of  the  parallel  rod.  Then  since  the  velocity  of  revolution  of  the 
circumference  of  the  wheel 
is  the  same  as  the  linear 
velocity  of  the  locomotive, 
it  follows  that  v=u.r/ri. 
Now  not  only  the  end  of  the 
parallel  rod,  but  every  point  B  B 

in  it,  is  revolving  with  the  Fi&-  165a 

velocity  v  in  a  circle  of  radius  r.  Thus  a  centrifugal  force  is 
generated  which  produces  flexural  stresses.  When  the  rod  is  at 
its  lowest  position  BB,  this  centrifugal  force  acts  as  a  downward 
uniform  load  producing  flexure;  at  the  highest  position  A  A  it  acts 
as  an  upward  uniform  load  producing  flexure;  at  the  position 
CC,  on  the  same  level  as  the  axles,  it  produces  a  compressive 
stress  in  the  direction  of  the  length  of  the  rod. 

Let  w  be  the  weight  of  the  parallel  rod  per  linear  unit;  then, 
from  rational  mechanics,  the  centrifugal  force  is, 

wf = wv2/gr  or  w'  =  wu2r/gri2 

which  may  be  called  the  centrifugal  load  per  unit  of  length. 
The  rod  being  a  beam  supported  at  its  ends,  having  a  length  /, 
a  breadth  b,  and  a  depth  d,  the  maximum  unit-stress  due  to 
this  uniform  load  is,  from  the  flexure  formula  (41), 


426 


MISCELLANEOUS  DISCUSSIONS 


CHAP.  XVIII 


which  is  the  flexural  stress  due  to  centrifugal  force  when  the 
bar  is  at  its  highest  or  at  its  lowest  position.  If  the  bar  is  not 
rectangular,  as  is  generally  the  case,  the  values  of  c  and  /  for  its 
cross -section  are  to  be  obtained  by  the  methods  which  are  ex- 
plained in  Arts.  42  and  43. 

In  this  formula  g  is  the  acceleration  of  gravity,  or  32.16  feet 
per  second  per  second.  In  using  it  in  formulas,  however,  all 
quantities  should  be  expressed  in  terms  of  the  same  linear  unit, 
the  inch  being  preferable.  For  example,  let  a  locomotive  be 
running  at  60  miles  per  hour,  the  radius  of  the  drivers  being 
3  feet  and  that  of  the  parallel  rod  i  foot,  this  being  of  steel,  4  inches 
deep,  2  inches  thick,  and  8  feet  long.  Here  u=SS  feet  per  second 
=  12X88  inches  per  second,  g  =  32. 16X12  inches  per  second 
per  cecond,  r\  =3X12  inches,  r  =  iXi2  inches,  /  =  8Xi2  inches, 
b  =  2  inches,  d=4  inches,  and  w  =  2. 2 7  pounds  per  linear  inch. 
The  centrifugal  load  per  inch  then  is  w' =61  pounds,  and  the 
maximum  flexural  stress  is  S  =  i$  200  pounds  per  square  inch, 
which  is  probably  not  sufficiently  low  when  it  is  considered  that 
the  parallel  rod  is  subject  to  rapid  alternating  stresses  and  per- 
haps to  shocks. 

The  connecting  rod  moves  in  a  circle  of  radius  r  at  the  crank 
pin,  while  the  other  end  moves  only  in  a  straight  line.  Thus 
at  the  end  A  there  is  no  centrifugal  load,  while  at  B  the  cen- 
trifugal load  is  the  same  as  given  by  the  above  expression  for 

wr .  When  the  rod  is  in  the 
position  shown  in  Fig.  1656,  it  is 
a  beam  acted  upon  by  a  cen- 
trifugal load  which  varies  uni- 
formly from  o  at  A  to  w'  at  B. 

The  total  load  is  hence  \wfl,  the 
Fig.  1650  .  .  ' 

reaction  at  A  is  Jw'/,  and  that  at 

B  is  \uu'l.  The  bending  moment  for  any  section  distant  x  from 
the  end  A  then  is, 

M=\w'l .  x—^wf(x2/l)  .  $x=%w'(l2x—x3)/l 

and  the  maximum  value  of  M  occurs  at  the  section  for  which 
which   gives   maximum    M=  0.0638 it/ 12.      Then   from 


ART.  166  UNSYMM ETHIC    LOADS   ON   BEAMS  427 

the  flexure  formula  (41),  the  flexural  unit-stress  is, 
S=Mc/I  S=  0.383  w'P/bd2 

the  second  being  for  a  rectangular  section  in  which  it/  has  the 
same  value  as  for  the  parallel  rod. 

By  comparing  the  formulas  for  rectangular  parallel  and 
connecting  parallel  rods  it  is  seen  that  the  unit-stress  5  for  the 
former  is  about  twice  as  great,  if  the  length  and  section  area  are 
the  came  in  the  two  cases.  The  parallel  rod  needs  the  greatest 
cross-section  at  the  middle,  while  the  connecting-rod  needs  the 
greatest  cross-section  at  about  o.6/  from  the  cross-head. 

Prob.  165.  The  connecting  rod  of  an  engine  is  2  feet  long  and  it 
is  attached  to  a  crank  pin  at  a  distance  of  6  inches  from  the  axis  of 
a  fly-wheel.  If  the  wheel  makes  750  revolutions  per  minute,  find  a 
square  cross-section  for  the  connecting  rod  so  that  the  centrifugal  unit- 
stress  5  may  be  4200  pounds  per  square  inch. 

ART.  166.     UNSYMMETRIC  LOADS  ON  BEAMS 

It  was  explained  in  Art.  43  that  two  axes  may  be  drawn  in 
the  plane  of  any  section  area  of  a  beam,  these  passing  through 
the  center  of  gravity  of  the  section  and  being  at  right  angles  to 
each  other,  so  that  the  moment  of  inertia  of  the  section  is  greater 
for  one  axis  and  less  for  the  second  axis  than  for  any  other  axis 
through  the  center  of  gravity.  These  are  called  'principal  axes  ', 
and  the  moments  of  inertia  with  respect  to  them  are  those  required 
in  all  common  cases;  Table  6  gives  these  moments  of  inertia 
for  I  sections  and  Table  8  those  for  T  sections.  The  great 
majority  of  beam  sections  are  symmetric  with  respect  to  both 
of  the  principal  axes,  while  a  few  like  the  T  and  bulb  sections 
are  symmetric  with  respect  to  one  principal  axis  only.  The 
L  section  of  Fig.  42c  and  the  Z  section  of  Fig.  42d  are  unsymmetric 
with  respect  to  both  principal  axes ;  these  are  not  commonly  used 
as  beams,  but  when  so  used,  the  flexure  formula  (41)  cannot 
be  applied  to  their  correct  discussion  without  the  determination 
of  the  moments  of  inertia  for  the  principal  axes. 

A  load  on  a  beam  is  said  to  be  'unsymmetric'  when  its  vertical 


428 


MISCELLANEOUS  DISCUSSIONS 


CHAP.  XVIII 


plane  does  not  coincide  with  one  of  the  principal  axes  of  the 
section  area.  The  simplest  case  is  that  shown  in  Fig.  166a, 
where  the  broken  lines  show  the  principal  axes  of  a  rectangular 
section,  and  the  load  P  is  applied  at  a  certain  distance  from  the 
vertical  axis.-  If  the  load  is  on  a  beam  with  supported  ends,  it 
is  plain  that  the  reactions  of  the  end  will  not  be  uniformly  dis- 
tributed over  the  supports  and  hence  the  right-hand  part  of  the 
section  will  be  but  slightly  stressed.  If  the  beam  is  fastened  to 
the  supports,  the  reaction  may  be  downward  or  negative  on  the 
right-hand  part  and  upward  on  the  left-hand  part,  so  that  tor- 
sional  stresses  will  be  developed.  This  case  is  one  that  should 
be  avoided,  for  it  is  clear  that  parts  of  the  beam  are  much  more 
highly  stressed  than  when  the  load  is  placed  so  that  its  resultant 
coincides  with  the  vertical  principal  axis. 


Fig.  166a  Fig.  1666  Fig.  166c  Fig.  IQQd 

Fig.  166£  shows  a  case  of  unsymmetric  loading  which  occurs  in 
a  purlin  beam  connecting  two  roof  trusses.  The  upper  chords  or 
main  rafters  of  the  truss  being  inclined  to  the  horizontal  at  the 
angle  <£,  the  upper  and  lower  surfaces  of  the  horizontal  purlin 
have  the  same  inclination,  and  the  load  on  its  upper  surface 
makes  the  angle  (f>  with  the  principal  axis  2-2.  Let  /i  and  1 2 
be  the  moments  of  inertia  of  the  section  area  with  respect  to  the 
principal  axes  i-i  and  2-2.  Let  M  be  the  bending  moment  at 
the  dangerous  section  due  to  the  vertical  loads,  then  the  compo- 
nent M  cos<£  acts  in  the  plane  2-2  and  produces  the  unit-stress 
Si  =c\M  cos$//i,  where  c\  is  the  distance  from  i-i  to  the  upper 
or  lower  surface  of  the  beam.  When  <£  is  not  a  large  angle,  the 
unit-stress  S\  is  frequently  computed  by  this  method  and  regarded 
as  the  actual  flexural  stress.  For  example,  let  the  angle  <£  be 
30  degrees,  the  simple  beam  be  4  inches  wide,  6  inches  deep, 
13  feet  in  span,  and  be  subject  to  a  total  uniform  load  of  600 
pounds.  Here  M  =  11  700  pound-inches,  /i=72  inches4,  £1=3 


ART.  166  UNSYMMETRIC    LOADS    ON    BEAMS  429 

inches,  and  then-  Si  =490  pounds  per  square  inch;   this  is  much 
smaller  than  the  actual  unit-stress. 

Another  method  of  dealing  with  the  last  case  is  to  consider 
the  neutral  axis  mn  as  horizontal,  to  find  the  values  of  c  and  / 
with  respect  to  it,  and  then  to  compute  the  unit-stress  from  the 
flexure  formula  S=Mc/I.  Here  c  is  the  distance  from  the 
remotest  fiber  at  the  upper  or  lower  corner  to  the  neutral  axis 
and  it  is  easy  to  show  that  c=c\  cos<£  +  £2  sin<£,  where  c\  and  c% 
are  the  coordinates  of  the  corner  with  respect  to  the  principal  axes 
i-i  and  2-2.  It  may  also  be  shown  that  /=/i  cos2<£-|-/2  sin2^ 
gives  the  moment  of  inertia  of  the  section  area  with  respect  to 
the  axis  mn.  For  the  above  numerical  example,  £1=3  and 
€2  =  2  inches,  /i=72  and  /2=32  inches4;  then  6=3.60  inches, 
7=62  inches4,  and  finally  S  =680  pounds  per  square  inch.  It 
is,  however,  not  correct  to  assume  that  the  neutral  axis  is  hori- 
zontal and  the  computed  value  of  S  is  too  small. 

A  more  precise  method  is  to  resolve  the  vertical  load  on  thfe 
purlin  into  components  parallel  to  the  principal  axes,  those 
normal  to  i-i  producing  the  moment  M  cos  96  and  those  normal 
to  2-2  producing  the  moment  M  sin<£.  Then  the  flexural  unit- 
stress  due  to  the  first  moment  is  Si  =c\M  cos<£//,  and  that  due 
to  the  second  moment  is  S2=c2M  sin<^//2.  The  actual  unit- 
stress  on  the  remotest  fiber  then  is, 


e       M  i  /i  A«\ 

02=^2    —  j—        —j  —  )  (166) 


For  the  above  numerical  example,  this  formula  gives  S  = 
pounds  per  square  inch,  which  is  a  more  reliable  value  than  those 
found  by  the  other  methods.  It  does  not,  however,  give  sufficient 
weight  to  the  influence  of  the  component  M  sin^>,  since  the  forces 
that  cause  this  moment  act  in  the  plane  of  the  upper  surface  of 
the  beam.  For  a  case  like  Fig.  166c,  where  the  forces  PI  and 
P2  act  in  the  lines  of  the  principal  axes,  this  method  is  strictly 
correct,  M  being  the  moment  due  to  the  resultant  P  which  makes 
the  angle  <£  with  2-2;  by  discussing  this  case  it  can  be  shown 
that  the  neutral  axis  mn  is  not  normal  to  the  plane  of  M  except 
when  1  1  and  J2  are  equal. 


430  MISCELLANEOUS  DISCUSSIONS  CHAP.  XVIII 

Formula  (166)  applies  to  the  discussion  of  the  Z  bar  shown 
in  Fig.  166d,  as  soon  as  the  values  of  c\,  c2,  /i,  1  2,  and  <£  are 
known.  Table  11  gives  data  for  a  few  Z  bars  with  equal  legs, 
this  being  a  part  of  the  table  in  the  Cambria  pocket  book.  The 
moments  of  inertia  Ia  and  /&  are  those  with  respect  to  the  axes 
a-a  and  b—b,  while  the  principal  axes  are  i-i  and  2-2.  The 
value  of  1  2  is  found  by  72=ar22,  where  a  is  the  section  area 
and  r2  the  least  radius  of  gyration.  The  value  of  I\  is  found 
by  subtracting  72  from  the  sum  /„  +  /&.  The  tangent  of  the 
angle  <£  is  given  in  the  sixth  column  of  the  table,  and  from  this 
the  values  of  cos<i  and  sin^  are  readily  found.  For  example, 
take  a  Z  bar  8  inches  deep  with  legs  3  inches  long,  and  let  a  load 
P  be  applied  as  shown  in  Fig.  166d  which  produces  a  bending 
moment  M  equal  to  120  ooo  pound-inches.  Making  a  drawing  of 
the  section  from  the  data  in  the  table,  the  measured  values  of 
c\  and  c2  normal  to  the  axes  i-i  and  2-2  are  found  to  be  4.61 
and  i.  60  inches.  From  the  given  value  tan<£=o.27,  there  are 
found  cos</>  =0.965  and  sin<£  =0.260.  Following  the  above  rules, 
the  principal  moments  of  inertia  are  72  =  2.56  and  71=50.24 
inches4.  The  formula  (166)  then  gives  5  =  30100  pounds  per 
square  inch  for  the  stress  on  the  upper  or  lower  corner  of  the 
end  of  the  leg.  By  using  the  rough  method  in  which  aa  is  taken 
as  the  neutral  axis,  there  is  found  5  =  120000X4/44.64  =  10800 
pounds  per  square  inch,  which  is  about  one-third  of  the  former 
value.  Consideration  of  Fig.  166a  indicates,  however,  that 
the  actual  unit-stress  in  the  corner  of  the  Z  bar  is  probably  higher 
than  30  100  pounds  per  square  inch. 

Prob.  166a.  Show  that  the  angle  (j)  which  gives  the  greatest  flexural 
unit-stress  for  the  cases  of  Fi.  1666  is  that  for  which 


Prob.  1666.  Show  that  the  angle  6  between  the  neutral  axis  mn 
and  the  principal  axis  i-i  in  Fig.  1666  is  given  by  tan0=  —  (7i/72)tan0. 
For  the  above  example  of  the  purlin  beam,  show  that  0  is  5  if  degrees, 
and  that  hence  the  neutral  axis  is  inclined  21}  to  the  horizontal. 

Prob.  166c.  Compute  the  unit-stress  5  for  a  Z  bar  6  inches  deep 
with  legs  3  J  inches  long,  when  used  as  beam  under  a  bending  moment 
of  90000  pound-inches. 


ART.  167 


INFLUENCE  LINES 


431 


ART.  167.    INFLUENCE  LINES 

An  influence  line  diagram  shows  the  variation  of  the  shear  or 
moment  at  a  given  section  of  a  beam  as  a  single  load  P  travels 
over  it,  the  values  of  the  shear  or  moment  at  the  given  section 
being  laid  off  as  ordinates  under  the  load.  Fig.  167a  gives  the 
shear  influence  line  for  the  section  K  of  a  simple  beam.  For  any 
load  P  on  the  beam  at  a  distance  id  from  A,  the  shear  at  K  is 
-f  P(i  —  K)  when  the  load  is  on  KB  and  —  PK  when  it  is  on  AK. 
Hence,  when  P  is  at  A  or  B  the  ordinate  is  o,  when  P  is  at  K 
the  ordinates  are  +P(i  —KI)  and  —  P/CI,  the  distance  AK  being  KI/, 
and  the  ordinates  on  each  side  of  K  vary  directly  as  the  distances 
from  A  and  B.  This  diagram  clearly  shows  that  every  load  on 


Fig.  167a 


Fig.  1676 


the  left  of  K  gives  a  negative  shear  at  K  and  that  every  load  on 
the  right  of  K  gives  a  positive  shear  at  K.  It  also  shows  that  the 
greatest  shears  at  K  occur  when  the  loads  are  near  K. 

Fig.  1676  gives  the  moment  influence  line  for  the  section  ^ 
of  a  simple  beam.  For  any  load  P  on  BK  the  bending  moment 
at  K  is  +Pl(i-K)K1  and  for  any  load  on  AK  it  is  +P//c(i-«i), 
where  K  and  KI  have  the  same  meaning  as  above.  Thus,  the 
moment  at  K  is  positive  and  it  varies  as  the  first  power  of  K  in  both 
cases,  and  hence  the  influence  diagram  consists  of  two  straight 
lines  which  are  readily  drawn  after  erecting  the  ordinate  P/(i  -KI)KI 
at  K.  This  diagram  shows  that  the  maximum  moment  at  K 
occurs  when  the  span  is  fully  loaded  and  when  the  heavy  loads 
are  near  K. 

Fig.  167c  gives  influence  lines  for  the  section  K  in  an  over- 
hanging beam,  the  upper  one  being  for  shear  and  the  lower  one 
for  bending  moment,  the  ordinates  showing  the  shear  or  moment 


432 


MISCELLANEOUS  DISCUSSIONS 


CHAP.  XVIII 


at  K  being  laid  off  under  the  moving  load  P.  These  lines  are 
readily  drawn  by  laying  off  as  ordinates  the  shear  and  moment 
at  K  for  a  load  at  K  and  then  drawing  straight  lines  through  the 
points  under  the  supports.  These  diagrams  show  that  the  greatest 
positive  shear  at  K  occurs  when  CA  and  KB  are  loaded,  that  the 
greatest  negative  shear  at  K  occurs  when  AK  and  BD  are  loaded, 
that  the  greatest  positive  moment  at  K  occurs  when  AB  is  loaded, 
and  that  the  greatest  negative  moment  at  K  occurs  when  CA  and 
BD  are  loaded. 

For  cantilever,  simple,  and  overhanging  beams,  the  influence 
lines  are  straight,  but  for  fixed  and  continuous  beams  they  are 


Fig.  167c 


Fig. 


curved.  Tig.  167  d  shows  the  diagrams  for  the  section  K  in  a 
continuous  beam  of  two  equal  spans,  the  upper  one  being  for  shear 
and  the  lower  one  for  moment.  Here  the  reactions  are  given 
by  Art.  73  and  for  a  kad  on  the  left  of  K  the  shear  at  K  is 
V=Ri—  P=  —  \P($K  —  K3)  from  which  a  few  ordinates  maybe 
computed  in  order  to  draw  the  curve;  also  for  a  load  on  the  span 
2-3  the  shear  at  K  is  V=Ri=  -JPC/c-Ac3).  Similarly  the 
moment  at  K  due  to  a  load  on  the  left  of  K  is  M  =  R\Kil  —  Pfal— /c/), 
where  K\l  is  the  distance  iK. 

Influence  lines  may  be  used  to  assist  in  finding  the  greatest 
positive  or  negative  values  of  the  shear  at  a  given  section  due 
to  a  uniform  load,,  For  this  purpose  P  is  to  be  replaced  by  w, 
the  load  per  linear  unit,  and  the  sum  of  all  the  shears  is  obtained 
from  the  areas  of  the  influence  diagrams.  Thus  the  greatest 


to  168  CURVED  BEAMS  433 

negative  shear  at  K  in  Fig.  167c  is  represented  by  the  areas  of 
the  triangles  in  the  first  diagram  below  AK  and  BD.  Let 
AK=Kil  and  BD=K2l,  where  /  is  the  span  AB.  Then  the 
negative  ordinates  for  K  and  D  are  PKI  and  PK2}  and  the  areas 
of  the  triangles  are  ^K^XWKI  and  ^K2lXwK2,  the  sum  of  which 
is  ^wl(Ki2  +  K22)-,  this  is  the  negative  shear  at  K  due  to  a  uniform 
load  on  both  AK  and  BD.  In  a  similar  way  the  greatest  positive 
or  negative  value  of  the  moment  at  a  given  section  due  to  a 
uniform  load  w  can  be  found, 

The  principal  value  of  influence  lines  is  in  representing  to 
the  eye,  more  clearly  than  formulas  can  do,  the  variation  of  the 
shear  or  moment  at  a  given  section,  thus  enabling  the  distribu- 
tion of  load  which  renders  the  shear  or  moment  a  maximum  to 
be  readily  determined.  The  lines  are  in  fact  graphic  represen- 
tatives of  the  algebraic  expressions  for  V  and  M  at  the  given  sec- 
tion due  to  a  load  which  travels  over  the  entire  length  of  the 
beam.  Their  greatest  value  is  in  the  analysis  of  partially  con- 
tinuous and  arch  bridges  (see  Roofs  and  Bridges,  Part  IV). 

Prob.  lC7a.  Draw  the  influence  line  for  the  reaction  at  A  in  the 
simple  beam  of  Fig.  167a, 

Prob.  1676.  Draw  the  influence  line  for  the  reaction  aCA  in  the 
overhanging  beam  of  Fig.  167c0 

Prob.  167c.  Draw  the  influence  line  for  the  moment  at  B  in  the 
overhanging  beam  of  Fig.  167c0 

ART.  168.    CURVED  BEAMS 

The  word  beam  usually  implies  a  bar  which  is  originally 
straight,  but  the  fundamental  assumptions  of  the  theory  of 
straight  beams  (Art.  39)  can  be  applied  to  the  deduction  of  for- 
mulas for  curved  pieces  which  are  subject  to  flexure.  The  most 
important  case  is  that  of  a  rectangular  section  where  the  upper 
and  lower  surfaces  are  curved  in  concentric  circles  as  seen  in 
Figs.  168a  and  1686,  Let  TI  and  r2  be  the  radii  of  the  concave 
and  convex  sides;  let  b  be  the  breadth  and  d  the  radial  depth, 
so  that  r2—ri  =  d.  Let  c?  and  c2  be  the  radial  distances  from 


434  MISCELLANEOUS  DISCUSSIONS         CHAP,  xvill 

the  neutral  surface  to  the  concave  and  convex  sides,  so  that 
Ci+c2  =  d.  It  will  be  shown  that  Ci  is  less  than  c2,  or  that  the 
neutral  surface  is  nearer  to  the  concave  than  to  the  convex  side, 
and  also  that  the  flexural  fiber  unit-stress  is  greater  on  the  con- 
cave than  on  the  convex  side. 

Assume,  as  in  Art.  40,  that  any  plane  section  Bb  which  is 
radial  before  flexure  remains  also  plane  after  the  flexure  and  takes 
the  position  Cc.  In  Fig.  IGSclet  Oa=Ob  =  r1  and  OA  =  OB  =  r2; 
also  let  non  be  the  neutral  surface  so  that  ob=ci  and  oB=c2o 


On  the  concave  side  the  fiber  ab  is  shortened  by  the  amount  be 
at  the  neutral  surface  the  fiber  no  remains  unchanged  in  length, 
and  on  the  convex  side  the  fiber  AB  is  elongated  by  the  amount 
BC.  Let  0  be  the  angle  between  the  radii  OA  and  OB  and  dO 
be  the  angle  between  Bb  and  Cc.  Then  ab=rld  and  bc  =  Cid6, 
so  that  the  unit  shortening  on  the  concave  side  is  CidO/riO;  also 
AB  =  r20  and  BC=c230,  so  that  the  unit  elongation  on  the  convex 
side  is  c2dd/r20.  Let  r  be  the  radius  of  the  neutral  surface  and  z 
be  any  radius  less  than  r2  and  greater  than  r\\  the  change  per 
unit  of  length  for  the  fiber  of  radius  z  is  then  pq/mp  or  (z  —r)d6/zd. 
Now  if  E  is  the  modulus  of  elasticity  of  the  material, 

36  Cl          9asF^£?          „       -30*-r 

Sl=E~d7i       S2=ET72       S=E*~JT 

are  the  fiber  unit-stresses  due  to  the  flexure,  Si  and  S2  being  those 
on  the  concave  and  convex  sides  of  the  beam  and  S  being  that 
for  any  radius  z.  For  a  curved  beam,  then,  the  fiber  stresses  are 
not  directly  proportional  to  their  distances  from  the  neutral 
surface.  Thus  while  laws  1-5  of  Arts.  39  and  40  are  valid  for 


ART.  168  CURVED  BEAMS  435 

curved  beams,  law  6  fails  and  the  neutral  surface  does  not  pass 
through  the  centers  of  gravity  of  the  cross-section.  While  Fig. 
168c  is  drawn  for  the  case  of  downward  curvature  and  positive 
bending  moment,  the  same  formulas  result  for  upward  curvature 
or  negative  bending  moment. 

The  position  of  the  neutral  surface  is  to  be  determined  by  the 
condition  that  the  algebraic  sum  of  all  the  stresses  normal  to 
the  radius  Bb  shall  be  zero,  or  if  da  is  the  elementary  area  bdz, 

§0     z-r      _       dd    CT\-r 
6        z          '        6    I        z      Z~° 

Jr\ 

Performing  the  integration  and  noting  that  r2  —TI  =  d}  there  results 
d— rlog  —  =o        or        r=d/log  — 

in  which  the  logarithms  are  Napierian.  Replacing  r  by  f 'i+Ci 
and  then  by  r2—c2,  there  are  found, 

ci  =  -n+d/log-  c2  =  r2-d/\ogT-  (168) 

TI  r\ 

which  give  the  radial  distances  from  the  neutral  surface  to  the 
concave  and  convex  sides  of  the  curved  beam. 

The  bending  moment  M  for  the  section  Bb  is  equal  to  the 
resisting  moment,  formed  by  the  algebraic  sum  of  the  moments 
of  all  the  stresses  normal  to  the  radius  Bb,  hence 


]    C 

7 

Jri 


Integrating  and  reducing  by  the  conditions  that  r2=r+C2,  Ti 
and  r=d/log  faAi)  there  is  found 


Replacbg  E  dd/0  by  its  values  from  (168)  in  terms  of  Si  and  S2, 

iMc\  „  2Mc2 

01=7-;  —  -,  -  r  ^2  =  73  —  1  -  \ 
c\)  bdr2(c2-ci) 


436 


MISCELLANEOUS  DISCUSSIONS 


CHAP,  XVIII 


which  are  the  unit-stresses  on  the  concave  and  convex  sides  of 
the  curved  beam  due  to  the  bending  moment  M . 

As  an  example,  let  ^2  =  36,  7*1  =  24,  and  d=i2  inches;  then 
log  (r2/ri)  =  log r2 -log ri  =  3.5835 -3. 1781  =  0.4054,  whence  by 
(168),  d  =  5.60  and  c2  =  6.40  inches.  Then  by  (168)'  Si  =  j.oM/bd2 
and  S2—$.$M/bd2.  For  a  curved  beam  the  fiber  unit-stress  on 
the  concave  side  is  always  greater  than  that  in  a  straight  beam. 
For  the  straight  beam  r<2  =  r1=cQ  and  it  can  be  shown  that  the 
above  formulas  reduce  to  Ci  =  C2=%d  and  Si  =  S2=6M/bd2. 

When  TI  is  very  small,  as  in  Fig.  168J,  then  Ci  is  very  small 
and  Si  is  very  large,  so  that  a  small  bending  moment  Pp  may 

cause  rupture.  Hence 
small  radii  should  be 
avoided  in  a  curved  piece 
which  is  subject  to 
flexure.  When  TI  =  O  ano1 
r2  =  d,  it  can  be  shown 


Ffc.  168d 


Fig.  W8e 


that  Ci  =  o,  c2  =  d,  Si  =  oo  , 


When  the  circles  are  not  concentric,  as  in  Fig.  W7e,  similar 
general  conclusions  result,  although  the  above  formulas  do  not 
directly  apply.  When  r2  is  less  than  r±  +  d,  the  following  approx- 
imate formulas  may  be  used,  in  which  h  is  the  distance  between 
the  centers  of  the  two  circles  and  d  is  the  depth  of  the  rec- 
tangular section  or  d=r2—ri+h, 

,        v7i  3     V^+h 

Ci=a 


Si  =  3  M/bdci  S2  =  3  A 

These  formulas  also  apply  approximately  to  the  case  of  -on- 
centric  circles  by  making  h=o.  Thus,  for  the  above  numerical 
example,  ^1  =  5.39  and  £2  =  6061  inches,  while  Si  =  6.'jM/bd2  and 


Prob.  168.   When  n  =  rt—  oo,   prove   that  formulas  (168)  reduce 
to  Ci  =  c*  =  \d.     Also  that  (168) '  reduce  to  S3  =  Sa  =  6M/bd*. 


ART.  169 


PRODUCT  OF  INERTIA 


437 


ART.  169o    PRODUCT  OF  INERTIA  • 

The  '  Product  of  Inertia  '  of  a  plane  area  with  respect  to 
two  rectangular  axes  is  the  quantity  Imxy,  which  is  the  sum  of 
the  products  formed  by  multiplying  each  elementary  area  m  by 
its  coordinates  x  and  y.  When  the  plane  area  is  symmetric  with 
respect  to  one  of  the  axes,  then  the  product  of  inertia  is  zero; 
thus  in  Figs.  169a  and  169&  the  axis  2-2  is  one  of  symmetry,  so 
that  for  any  assigned  value  of  y  there  are  two  equal  values  of  x 
with  opposite  signs;  also  in  Fig.  169c  the  axis  i-i  is  one  of 
symmetry  and  Imxy  is  zero,  because  for  any  assigned  x  there  are 


i 


— i 


— i 


i — 


Fig.  169a        Fig.  1696 


Fig.  169c        Fig. 


Fig.  169e 


two  equal  values  of  y  with  opposite  sign.     For  Figs.  169J  and  169e 
aeither  axis  is  one  of  symmetry  and  Imxy  is  not  zero. 

For  a  plane  surface  which  is  symmetric  with  respect  to  an 
axis  through  its  center  of  gravity,  the  product  of  inertia  with  respect 
to  another  set  of  rectangular  axes  parallel  to  the  first  set  is  equal 


&~ 

-*-t-^ 

sq-f 

—2 

V      . 

ffi 

V 

-M'-- 
J  i\ 

'       2 


Fig.  169/     Fig.  169g 


Fig.  169& 


Fig.  169* 


to  the  area  of  the  surface  multiplied  by  the  coordinates  of  its 
center  of  gravity.  Let  X'-X'  and  Y'-Y'  (Fig.  169&)  be  tfie  two 
rectangular  axes  through  the  center  of  gravity,  and  let  y  and  x 
be  the  coordinates  of  that  center  of  gravity  with  respect  to  the 
axes  X-X  and  Y-Y.  Let  x'  and  /  be  the  coordinates  of  any 


438  MISCELLANEOUS  DISCUSSIONS  CHAP,  xvili 

element  m  with  respect  to  the  first  set  of  axes  and  x  and  y  the 
coordinates  with  respect  to  the  second  set*  Then  x=x+xf  and 
y=~y+y',  and 

Imxy  =  xylm  +xZmyf+ySmx'  -f  Imx'y' 

But  'Imyf  and  2mx'a,re  each  zero  from  the  definition  of  centei 
of  gravity  and  Imxfyf  is  zero  because  the  area  has  an  axis  of 
symmetry.  Hence 

Imxy  =  xylm  =  axy 

which  proves  the  proposition  stated.  For  example,  take  the  angle 
section  in  Fig.  169^  the  dimensions  being  5X3X2  inches,  and 
its  center  of  gravity  being  0.75  inch  from  the  back  of  the  longer 
leg  and  1.75  inches  from  the  back  of  the  shorter  leg.  This 
section  can  be  divided  into  two  rectangles,  as  shown,  and  the 
coordinates  of  their  centers  of  gravity  be  readily  found,  then 

jr««y-(SXo.s)(o.75J(o.s)+(2.5Xo.s)(-i.s)(-i.o)-+2.8i2S  ins4 

is  the  product  of  inertia  for  the  angle  section  with  respect  to 
the  axes  X-X  and  Y-Y. 

Products  of  inertia,  unlike  moments  of  inertia,  may  be  either 
positive  or  negative.  Thus  the  product  of  inertia  for  Fig.  169d 
is  positive  while  that  for  Fig.  169e  is  negative.  For  a  rectangle 
or  isosceles  triangle  axes  can  always  be  drawn  so  that  Imxy  is 
zero.  For  the  right-angled  triangle  in  Fig.  169/  the  product  of 
inertia  for  the  axes  X-X  and  Y-Y  is 


n 


while  for  the  parallel  axes  through  the  center  of  gravity,  as  in 
Fig.  169g,  the  theorem  of  the  last  paragraph  shows  that  the 
product  of  inertia  is  +TV&2d2° 

Prob.  169.  Find  the  product  of  inertia  for  a  rectangle  for  two 
axes  through  its  center  of  gravity,  one  axis  being  a  diagonal  of  the 
rectangle. 

ART.  170.    MOMENT  OF  INERTIA 

The  term  '  Moment  of  Inertia  '  originated  from  the  theory  of 
a  rotating  body.  When  a  body  is  put  into  rotary  motion  around 


ART.  170  MOMENT  OF  INERTIA  439 

an  axis,  each  particle  offers  a  resisting  force  by  virtue  of  its  inertia, 
this  force  being  proportional  to  the  mass  m  of  the  particle  and  to 
its  acceleration.  Let  a  be  the  acceleration  of  a  particle  at  the 
distance  unity  from  the  axis,  then  the  acceleration  of  a  particle 
at  the  distance  z  is  za.  The  resisting  force  of  inertia  of  that 
particle  is  mza  and  the  moment  of  that  force  with  respect  to  the 
axis  is  mz2a.  The  expression  Imz2a  or  almz2  is  the  sum  of  all 
these  resisting  moments  when  the  summation-  is  extended  to 
cover  the  entire  rotating  body.  Now  the  quantity  Imz2  was 
called  the  moment  of  inertia  of  the  body  in  the  early  days  of 
mechanical  science  when  the  subject  of  inertia  was  imperfectly 
understood.  This  name  has  continued  in  use,  although  now  it 
is  clearly  understood  that  2mz2  is  not  the  sum  of  the  moments 
of  all  the  forces  of  inertia,  but  only  proportional  to  that  sum, 
the  angular  acceleration  a  having  been  omitted. 

Strictly  speaking  a  plane  surface  can  have  no  inertia  and 
hence  no  moment  of  inertia,  but  the  quantity  2mz2  is  called 
moment  of  inertia  when  m  is  any  elementary  area  and  z  is  its 
distance  with  respect  to  an  assumed  axis.  This  quantity  occurs 
so  frequently  in  mechanics  that  it  is  necessary  to  have  a  name 
for  it,  and  c  Moment  of  Inertia'  is  in  universal  use.  Hence, 

The  moment  of  inertia  of  a  plane  surface  with  respect 
to  an  axis  is  obtained  by  multiplying  each  elementary  area 
by  the  square  of  its  distance  from  that  axis  and  then  taking 
the  sum  of  all  these  products. 

The  word  *  axis '  used  above  means  a  straight  line  drawn 
usually  in  the  plane  of  the  surface  or  sometimes  drawn  normal 
to  that  plane.  When  the  term  moment  of  inertia  is  used  without 
qualification,  the  axis  is  understood  to  be  in  the  plane  of  the 
surface.  When  the  axis  is  normal  to  the  plane  the  quantity 
Imz2  is  called  the  c  Polar  Moment  of  Inertia.'  It  is  proved  in 
Art.  90  that  the  polar  moment  of  inertia  is  equal  to  the  sum  of 
the  moments  of  inertia  about  two  rectangular  axes  in  the  plane 
of  the  surface.  The  present  article  relates  only  to  moments  of 
inertia  about  such  rectangular  axes. 


440 


MISCELLANEOUS  DISCUSSIONS 


CHAP.  XVIII 


Moments  of  inertia  of  regular  geometric  surfaces  are  deduced 
by  help  of  the  calculus  in  the  manner  indicated  in  Art.  43.  The 
moment  of  inertia  Imx2  is  less  for  an  axis  passing  through  the 
center  of  gravity  of  the  surface  than  for  any  parallel  axis,  since 
Imx2  is  a  minimum  when  2mx  =  o,  and  the  latter  is  the  condi- 
tion that  the  axis  passes  through  the  center  of  gravity.  Let  / 
be  the  moment  of  inertia  2mx2  for  the  axis  through  the  center 
of  gravity,  I\  that  for  any  parallel  axis,  and  h  the  distance  between 
the  two  axes  (Fig.  170a).  Then, 


7i  =  2m 
Here  Imx  equals  o,  and  Im  is  the  entire  area  a,  hence 

h=I+ah2        and        I  =  Tl-ah2 

are  the  practical  rules  for  transferring  moments  of  inertia  from 
and  to  an  axis  through  the  center  of  gravity. 

The  principal  axes  of  a  plane  surface  are  those  which  pass 
through  its  center  of  gravity  in  such  directions  that  the  moment 


x— 


Fig.  170a 


Fig.  170c 


of  inertia  with  respect  to  one  axis  has  the  greatest  possible  value, 
while  that  with  respect  to  the  other  axis  has  its  least  possible 
value.  These  moments  of  inertia  are  generally  designated  by 
/i  and  L2  and  are  called  the  principal  moments  of  inertia.  For 
a  rectangle  with  sides  b  and  d}  where  d  is  greater  than  b,  the  prin- 
cipal moments  of  inertia  are  J1  =  T\M3  and  I2^=^b3d.  The 
principal  axes  are  always  at  right  angles  to  each  other. 

When  the  moments  of  inertia  with  respect  to  two  rectangular 
axes  X-X  and  Y-Y  are  known  and  also  the  product  of  inertia 
Imocy  (Art.  169)  with  respect  to  those  axes,  the  principal  moments 
of  inertia  may  be  found  as  follows:  Let  x  and  y  be  the  coordi- 


ART.  170  MOMENT  OF  INERTIA  441 

nates  cf  any  elementary  area  m  with  respect  to  the  axes  X-X 
and  Y-Y  (Fig.  1706)  and  let  x±  and  y^  be  the  coordinates  with 
respect  to  the  principal  axes  i-i  and  2-2,  Let  6  be  the  angle 
which  i-i  makes  with  X-X.  Then 

xi  =y  cos  0—  x  sin  0  yi  =y  sin  6-\-x  cos  6 

Squaring,  multiplying  by  m,  taking  the  sums,  replacing  Imx-f 
by  /2,  I  my  i2  by  /i,  I  my2  by  /*,  and  Imx2  by  /„,  there  results 

1  2  =Iz  cos2  0+/v  sin2  0  —  (Imxy)  sin  20 


Ii  =  Ix  sin2  0+7,  cos2  6  +  (2mxy)  sin  20 

Differentiating  each   of  these  with   respect  to  0   and   equating 
the  derivatives  to  zero,  there  is  found  for  each, 


tan  20  =  2(2mxy)/(Iy-Ix]  (170)' 

as  the  condition  which  renders  /i  and  1  2  a  maximum  or  a  mini- 
mum, the  same  condition  applying  to  both  because  tan  26  belongs 
to  two  angles  which  differ  by  90°.  Substituting  the  value  of  0 
from  (170)'  in  (170),  there  results 


72  or  h  =J(/*+/,,)±V \(Iy-Itf  +  (Smxy}*          (170)" 

which  gives  the  two  principal  moments  of  inertia,  the  largest 
resulting  from  the  +  sign  before  the  radical  and  the  smallest 
from  the  minus  sign.  One  of  these  values  is  1 2  and  the  other  is  7i» 

For  example,  let  Fig.  169*  represent  a  5X3X1  inch  angle 
section,  the  area  a  being  3.75  square  inches,  the  two  axes  X-X 
and  Y-Y  passing  through  its  center  of  gravity  and  the  values 
of  Ix  and  Iy  being  2.58  and  9.45  inches4.  Here,  as  shown  in 
Art.  169,  the  product  of  inertia  Imxy  is  +2.8125.  Then  (170)" 
gives  1 2=  10.45  and  I\=  1-58  inches4.  Also  (170)'  gives  tan  2  0  = 
0.8421,  from  which  0=  19°  39'  or  tan  $  =  0.357.  The  least  radius 
of  gyration  of  the  section,  found  from  r2=7j/#,  is  ^=0.65  inches. 

Adding  together  the  two  equations  in  (170)  there  results 
7i+72  =  7z  +  7J/,  that  is,  the  sum  of  the  principal  moments  of 
inertia  is  equal  to  the  sum  of  the  moments  of  inertia  with  respect 
to  any  two  rectangular  axes  through  the  center  of  gravity. 


442  MISCELLANEOUS  DISCUSSIONS  CHAP,  xvm 

When  1  1  and  1  2  are  known  the  moment  of  inertia  Ix  with 
respect  to  an  axis  X-X  which  makes  an  angle  <j>  with  the  axis 
i-i  is  readily  found  as  follows:  From  Fig.  170c  the  ordinate  is 
y'  —  y  cos  <f>—x  sin  <£.  Squaring,  multiplying  by  mi  and  replacing 
the  sums  2myf,  I  my,  Imx,  by  Ix,  I\,  1  2,  there  is  found 


When  either  of  the  axes  i-i  or  2-2  is  an  axis  of  symmetry,  Imxy 
equals  zero  (Art.  169),  and  then 


Ix  =  1  1  cos2  (/>  +/2  sin2  <f> 

When  /i  and  /2  are  equal,  each  being  called  /,  then  /*  =  /,  that 
is,  when  a  plane  surface  has  its  two  principal  moments  of  inertia 
equal  to  /,  and  one  of  them  is  an  axis  of  symmetry,  then  the 
moment  of  inertia  with  respect  to  any  axis  through  the  center 
of  gravity.  is  also  equal  to  /.  Hence  the  moments  of  inertia  of  a 
square  are  equal  for  all  axes  through  its  center  of  gravity,  and 
the  same  is  the  case  for  sections  like  Figs.  76a,  766,  76c,  when 
/i  and  72  are  equal. 

Prob.  170a.    Deduce  (170)"  from  (170)  and  (170)'. 

Prob.  1706.  For  the  Z  bar  of  Fig.  166J,  let  depth  of  web  be  6| 
inches,  width  of  flanges  3|  inches,  and  thickness  of  both  J  inch. 
Compute  the  principal  moments  of  inertia. 

ART.  171.    SPRINGS 

A  cantilever  beam  formed  of  two  or  more  superimposed 
plates,  as  in  Fig.  17  la,  may  be  used  as  a  spring.  Each  of  the 
plates  acts  separately  as  a  beam,  so  that  the  unit-stress  at  the 
dangerous  section  is  S=6Pl/nbd2,  where  P  is  the  load  at  the 
end  of  the  cantilever,  /  its  length,  b  its  constant  breadth,  d  the 
depth  of  each  plate,  and  n  the  number  of  plates.  Hence  when  P 
is  given  and  S  is  taken  at  the  safe  allowable  value,  the  number  of 
plates  required  is  n=6Pl/Sbd2.  The  deflection,  as  shown  in 
Art.  58,  for  a  triangular  beam  of  constant  depth  and  a  triangular 
plan  of  greatest  width  nb,  is  /=  6PP/Enbd3,  or  /=  SI2  /Ed.  Hence 
this  formula  applies  exactly  to  the  spring  seen  in  elevation  and 


ART.  171 


SPRINGS 


443 


plan  in  the  upper  part  of  Fig.  17  la,  the  ends  of  the  n  plates 
being  triangular  as  shown,  n  being  an  even  number,  and  very 
closely  to  the  spring  having  n  plates  with  rectangular  ends  as 


Fig.  171a 


Fig.  1716 


seen  in  Fig.  1716.  All  these  formulas  are  valid  only  when  the 
greatest  unit-stress  S  does  not  exceed  the  elastic  limit  of  the 
material. 

The  double  spring  of  Fig.  17 Ib  supports  a  part  of  the  weight 
of  a  wagon  body  at  A,  this  load  being  transferred  at  the  ends 
to  the  lower  spring,  by  which  it  is  carried  down  to  the  axle  at  B. 
2P  being  the  total  load,  P  the  reactions  at  the  ends,  and  /  the  half- 
length,  the  formulas  above  given  apply  also  to  this  case.  For 
example,  let  2P  be  600  pounds,  /  be  15  inches,  b  be  2  inches,  and 
let  each  spring  have  six  plates  of  ^-inch  thickness.  Then  S= 
6  X  300  X  1 5/6  X  2  X  (J)2  =36  ooo  pounds  per  square  inch,  which  is 
somewhat  more  than  one-half  of  the  elastic  limit  of  spring  steel. 
Also/=  36  ooo X  i52/3°  °°°  ooo X^=  i.i  inches.  The  work  stored 
in  the  double  spring  then  is  4(|P/)  =  66o  inch-pounds.  When  the 
plates  are  so  firmly  fastened  together  that  the  entire  section  acts 
like  a  beam,  both  the  unit-stress  and  the  deflection  are  much  de- 
creased. For  this  case  S=66Pl/b(nd)2  and  f=Sl2/En2d;  thus, 
for  the  above  numerical  data,  5=9  ooo  pounds  per  square  inch 
and  f=  0.03  inches. 

A  spiral  spring  fastened  at  one  end  to  a  fixed  axis  has  the 
other  end  attached  to  the  circumference  of  a  cylindrical  box 
which  is  turned  around  the  axis  by  the  spring  after  it  is  wound. 


444  MISCELLANEOUS  DISCUSSIONS  CHAP.  XVIII 

If  r  is  the  radius  of  the  box  (Fig.  171c),  S  the  unit-stress  in  the 
spring  at  the  point  where  it  is  connected  to  the  axis,  b  and  d 
the  thickness  and  width  of  the  rectangular  spring,  then  the 


Fig.  171c  Fig.  171e 

twisting  force  exerted  tangentially  at  the  circumference  of  the  box 


A  helical  spring  is  formed  by  winding  a  wire  around  a  cylinder 
which  is  then  removed.  In  Fig.  17  Id  let  the  axial  tensile  load  P 
be  applied  at  the  center  of  the  coil  and  let  D  be  the  diameter  of 
the  coil.  Then  the  twisting  moment  Px^D  produces  torsion 
throughout  the  wire.  Let  d  be  the  diameter  of  the  wire  and  Ss 
the  shearing  unit-stress  produced  at  the  circumference  of  the 
section  \nd2.  From  Art.  92  this  unit-stress  is  Ss=%PD/xd3)  or 
the  load  for  a  given  allowable  Ss  is  P  =  xd3Ss/8D.  The  distance 
through  which  the  load  moves  being  /,  the  work  performed  on 
the  spring  is  \Pf  and  the  equal  internal  energy  stored  in  the  wire 
is  \(Ss2/F)al  (Art.  122),  where  F  is  the  shearing  modulus  of 
elasticity  of  the  wire  material,  a  is  the  section  area  ^d2  and  / 
is  the  length  of  the  wire  in  the  coils  of  the  spring.  Equating 
these  two  expressions,  there  is  found  /=  SsDl/Fd  for  the  deflec- 
tion /.  When  there  are  n  coils  in  the  spring,  the  length  /  is 
approximately  nnD,  or,  if  h  is  the  length  of  the  spring  parallel 
to  its  axis,  then  I  is  given  exactly  by  I2=h2  +  (nnD)2. 

Another  method  of  deducing  the  above  formulas  is  to  imagine 
the  wire  stretched  out  in  the  straight  line  of  length  /  and  be 
subject  to  the  twisting  moment  Pp  (Fig.  17  le).  Then  by 


ART.  171  SPRINGS  445 

Art.  92,  Ss=  i6Pp/Kd*.  Also  when  P  moves  through  the  distance 
/  a  point  on  the  surface  of  the  free  end  of  the  wire  moves  through 
the  distance  fX^d/p  and  hence  the  angular  deformation  per 
unit  of  length  \sfd/2pl.  Then  the  modulus  of  elasticity  is  F=Ss/e 
(Art.  93),  or  F=SsX2pl/fd.  Replacing  p  by  the  actual  lever 
arm  JZ>,  there  are  found  for  Ss  and /the  formulas  above  given. 

As  an  example,  let  steel  wire  ^  inch  in  diameter  be  closely 
wound  around  a  cylinder  ^&  inch  in  diameter,  thus  forming 
a  helical  spring  2\  inches  long  and  having  36  coils.  Here 
P  =  ^  +  ^g-=J  inch.  For  a  shearing  unit-stress  of  30  ooo  pounds 
per  square  inch,  the  above  formula  gives  P=$.J  pounds  as  the 
greatest  allowable  tensile  load,  and  this  must  be  slowly  and 
gradually  applied.  The  elongation  of  the  spring  under  this  load, 
taking  the  shearing  modulus  of  elasticity  F  as  12  ooo  ooo  pounds 
per  square  inch,  is  /=  1.131  inches  for  the  approximate  value  of  / 
and /=  1. 132  inches  for  the  exact  value. 

The  above  formulas  apply  also  to  an  axial  compressive  force 
P  acting  on  a  helical  spring,  provided  it  be  not  too  long  and  pro- 
vided that  the  coils  are  sufficiently  open  to  allow  of  axial  shorten- 
ing. The  unit-stress  S*  due  to  the  load  P  must  be  always  less 
than  the  elastic  limit  of  the  material. 

The  energy  which  can  be  stored  in  springs  is  small  in  pro- 
portion to  their  weight.  Thus,  for  a  helical  spring  under  a 
tensile  load  (Fig.  17  Id)  the  stored  work  is  \(Ss2/F)al.  If  this 
stored  work  is  550  foot-pounds,  then  for  5=30000  and  F=. 
ii  200  ooo  pounds  per  square  inch,  the  volume  al  is  82  cubic 
inches,  so  that  the  weight  of  the  spring  must  be  240  pounds.  If 
this  work  is  expended  in  one  second,  one  horse-power  is  developed; 
if  in  one  minute  only  one-sixtieth  of  a  horse-power  is  developed. 
On  account  of  the  heavy  weight  required,  metallic  springs  are 
rarely  used  for  the  storage  of  energy  except  in  timepieces,  music 
boxes  and  mechanical  toys.  Their  principal  uses  are  in  weigh- 
ing machines,  as  in  spring  balances  and  dynamometers,  and  for 
absorbing  the  energy  due  to  sudden  forces  and  shocks,  as  in  the 
springs  of  wagons  and  cars. 


446  MISCELLANEOUS  DISCUSSIONS          CHAP.  XVIII 

Prob.  1710.  Show  that  the  distance  which  P  moves  in  Fig.  171d 
in  causing  the  unit-shear  S  in  the  wire  is  f=  2rlS/dE. 

Prob.  1716.  Show  that  the  work  stored  in  the  helical  spring  of 
Fig.  1715  is  sufficient  to  raise  a  load  W  equal  to  the  weight  of  the 
spring  through  a  height  of  4.9  feet,  when  5  =  30  ooo  and  through  a 
height  of  8.6  feet  when  5  =  40  ooo  pounds  per  square  inch. 

Prob.  171c.  Show  for  the  helical  spring  that  the  deflection  f  is 
proportional  to  the  axial  load  P  provided  the  wire  of  the  spring  is 
not  stressed  above  its  shearing  elastic  limit. 


ART.  172  INTRODUCTION  447 


CHAPTER   XIX 
MATHEMATICAL  THEORY  OF  ELASTICITY 

ART.  172.     INTRODUCTION 

The  mathematical  theory  of  elasticity  is  that  science  which 
treats  of  the  behavior  of  bodies  under  stress  when  the  law  of 
proportionality  of  deformation  to  stress  is  observed.  All  the 
theoretic  formulas  of  the  preceding  pages  have  been  derived 
by  the  help  of  this  law,  but  these  constitute  only  a  part  of  the 
mathematical  theory  of  elasticity.  The  formulas  derived  for  the 
deformation  of  bodies  under  tension  or  compression  suppose  the 
bodies  to  be  homogeneous  or  isotropic,  so  that  the  modulus  of 
elasticity  E  is  the  same  for  all  directions;  some  materials,  how- 
ever, have  different  properties  of  stiffness  in  different  directions 
so  that  there  may  be  several  values  of  E  to  be  considered,  this 
being  especially  the  case  with  crystals.  The  theory  of  elasticity 
takes  account  of  such  non-homogeneous  structure  and  deduces 
formulas  for  the  deformations  due  to  forces  applied  in  different 
directions.  In  this  chapter  only  the  elements  of  this  theory  can 
be  given,  and  in  general  the  bodies  under  stress  will  be  regarded 
as  homogeneous.  The  complete  theory  deals  not  only  with 
elastic  solids,  but  with  fluids,  gases,  and  the  ether  of  space,  while 
the  discussion  of  stresses  and  deformations,  both  in  homogeneous 
and  crystalline  bodies,  leads  to  the  investigation  of  wave  propa- 
gations, the  time  and  velocity  of  elastic  oscillations,  and  numerous 
other  phenomena  of  physics. 

Statics  proper  is  concerned  only  with  rigid  bodies,  while  the 
theory  of  elasticity  deals  with  bodies  deformed  under  the  action 
of  exterior  forces  and  which  recover  their  original  shape  on  the 
removal  of  these  forces.  All  the  principles  and  methods  of 
statics  apply  in  the  discussion  of  elastic  bodies,  but  in  addition 
new  principles  based  upon  Hooke's  law  arise.  The  amount  of 
deformation  being  small  within  the  elastic  limit  for  common 


448  MATHEMATICAL  THEORY  OF  ELASTICITY     CHAP.  XIX 

materials,  it  is  allowable  to  neglect  the  squares  and  higher  powers 
of  a  unit- elongation  in  comparison  with  the  elongation  itself, 
as  set  forth  in  Art.  13.  By  the  help  of  this  principle,  the  elas- 
tic change  in  volume  of  a  body  may  be  found  (Art.  173),  the 
modulus  of  elasticity  for  tension  or  compression  is  found  to  have 
a  certain  relation  to  the  modulus  of  elasticity  for  shearing  (Art. 
181),  and  a  new  modulus  of  elasticity  based  on  change  in  volume 
is  introduced  (Art.  182). 

The  general  case  of  a  body  acted  upon  by  forces  in  several 
directions  occupies  the  main  part  of  this  chapter,  this  case  requir- 
ing the  use  of  three  rectangular  coordinates.  This  chapter,  then, 
is  an  extension  of  the  methods  of  Chapters  XI  and  XV,  and 
includes  those  methods  as  special  cases.  It  has  been  prepared 
from  the  point  of  view  of  the  engineer  rather  than  that  of  the 
pure  mathematician,  and  should  be  regarded  only  as  an  intro- 
duction to  the  mathematical  theory  of  elasticity. 

The  student  should  consult  the  article  on  Elasticity  by  Kelvin 
in  the  Encyclopaedia  Britannica,  as  also  the  History  of  Todhunter 
and  Pierson.  The  works  of  Clebsch  (Elasticitat  fester  Korper, 
1862),  Winkler  (Elasticitat  und  Festigkeit,  1867),  Grashof  (Theorie 
der  Elasticitat  und  Festigkeit,  1878),  and  Flamant  (Resistance  des 
Materiaux,  1886)  may  be  mentioned  as  treating  the  subject  both 
from  the  theoretical  and  the  engineering  point  of  view. 

Prob.  172.  Consult  Todhunter  and  Pierson's  History  of  the  Theory 
of  Elasticity  and  of  the  Strength  of  Materials,  and  ascertain  some- 
thing about  the  important  investigations  of  Saint  Venant. 

ART.  173.     ELASTIC  CHANGES  IN  VOLUME 

The  changes  in  section  area  and  in  volume  which  occur  when 
a  bar  is  under  axial  stress  have  been  discussed  in  Art.  13,  and 
the  same  method  will  now  be  applied  to  the  case  of  a  body  acted 
upon  by  forces  in  different  directions.  When  the  elastic  limit 
of  the  material  is  not  surpassed,  the  deformation  due  to  any 
applied  force  is  proportional  to  that  force  for  deformations  in 
any  and  all  directions,  and  the  principles  of  Art.  139  enable 
these  to  be  determined. 


ART.  173  ELASTIC   CHANGES   IN   VOLUME  449 

The  simplest  case  is  that  shown  in  Fig.  173,  where  a  parallele- 
piped is  acted  upon  by  the  tensile  unit-forces  Si,  £2,  6*3  in  three 
rectangular  directions.  The  body  is  regarded  as  homogeneous, 
so  that  the  factor  of  lateral  contraction  A  and  the  modulus  of 
eksticity  E  are  the  same  in  all  directions.  Let  e\  be  the  unit- 
elongation  due  to  Si  or  ei=Si/E,  also  £2  =  $2/E  and  £3=Ss/E. 
Then  the  actual  unit- elongations  which  take  place  in  the  three 
rectangular  directions  are  given  by, 

e'  =  ei  —  yU2— >^3  £"=-£2— ^£3— M  £"'=£3— A£I  —  ^£2 

Now  for  any  cube  of  edge  unity,  the  volume  after  the  application 
of  the  unit-forces  Si,  S2,  S3  is  (i  +  £r)(i  +  e")(i  +  fi"'),  and  since 
ef,  e",  e"'  are  small  compared  with  unity,  this  product  is 
i  +  er  +  e"  +  e'"  (Art.  13).  Accordingly,  the  elastic  change  in 
the  unit  of  volume  is, 

s>+s»+£<»==:(I-2X)(£1  +  s2+e3)~(i-2X)(Si+S2+S8)/E  (173) 
Here  it  is  seen  that  there  is  no  change  in  volume  when  A  =  J. 
For  all  the  materials  of  construction  it  is  found  that  the  factor 
of  lateral  contraction  is  less  than  J,  so  that  they  increase  in  volume 
under  tension  arid  decrease  under  compression. 

.     In  the  above  discussion  the  unit-forces  Si,  S2,  S3  are  regarded 

as  tensile,  but  the  formula  for  change  of  volume  applies  equally 

well  when  one  or  all  of  them  are  compressive;    for  example,  if 

Si  is  tension  and  S2  and  S%  are  compression,  the  values  of  S% 

and  S3  are  to  be  taken  as  negative.     The  above  formula  refers 

to  unit  of  volume  and  the  actual  change  in  a  parallelepiped  of 

given  dimensions  is  found  by  multiplying 

the  unit-change  by  the  number  of  units 

of  volume  in  the  body.     For  example,  let 

a  brick  2X4X8  inches  be  subjected  to  a 

compression  of  12  800  pounds  upon  the 

two  flat  sides,  to  a  compression  of   4800 

pounds  upon  the  two  narrow  sides,  and 

to  a  tension  of   i  600   pounds    upon   the  two  ends. 

- 12800/32  =  -400,  S3  =  -4800/16  =  -300,    and   Si  =  +  1 600/8 

=  +200    pounds    per    square    inch;    then    Si +£2 +  £3=  —  500 

pounds    per   square    inch.     Taking    A  =  o.2,    and    £  =  2000000 


450  MATHEMATICAL  THEORY  OF  ELASTICITY     CHAP.  XIX 

pounds  per  square  inch,  the  change  per  unit  of  volume  is  found 
to  be  —0.00015,  so  tnat  the  decrease  in  the  volume  of  the 
parallelepiped  is  0.0002X64  =  0.0096  cubic  inches. 

Elastic  changes  in  section  area  due  to  the  action  of  several 
forces  are  readily  expressed  in  a  similar  manner.  For  instance, 
in  the  unit  cube  let  a  plane  be  drawn  normal  to  Si  cutting  out 
a  square;  its  area  after  the  application  of  the  three  forces  is 
(i  +  e")(i +  e'"),  which  is  equal  to  i-fV'  +  e'".  Hence  the 
change  per  unit  of  section  area  is  found  from  (173)  by  making 
e'  and  Si  equal  to  zero.  Thus,  for  the  above  numerical  example, 
the  change  per  unit  of  section  area  normal  to  Si  is  —0.00004, 
so  that  the  decrease  in  this  section  area  is  0.00004X8  =  0.00032 
square  inches. 

When  a  body  is  under  uniform  compression  in  all  directions, 
as  is  the  case  when  it  is  subjected  to  fluid  pressure,  the  unit- 
stresses  Siy  S2j  S$  are  equal,  as  also  the  unit-deformations  ei, 
£2,  £3.  For  this  case  the  change  per  unit  of  length  is  the  same 
in  all  directions  and  equal  to  (i—  2^)5,  the  change  per  unit  of 
area  in  any  section  is  2(1  —  2 A)  £,  and  the  change  per  unit  of  volume 
is  3(1  —  2A)e,  where  e  is  the  unit-change  S/E  which  would  be 
caused  by  an  axial  unit-stress  S  which  is  equal  to  the  uniform 
compressive  unit-stress.  The  change  in  section  area  is  hence 
double,  and  the  change  in  volume  is  three  times  that  in  a  linear 
dimension. 

Prob.  173.  Make  experiments  upon  india  rubber  with  the  inten- 
tion of  finding  the  value  of  the  factor  of  lateral  contraction  for  that 
material. 

ART.  174.    NORMAL  AND  TANGENTIAL  STRESSES 

The  general  case  of  internal  stress  is  that  of  an  elementary 
parallelepiped  held  in  equilibrium  by  apparent  stresses  applied 
to  its  faces  in  directions  not  normal.  Here  each  oblique  stress 
may  be  decomposed  into  three  components  parallel  respectively 
to  three  coordinate  axes,  OX,  OY,  OZ.  Upon  each  of  the  faces 
perpendicular  to  OX  the  normal  component  of  the  oblique  unit- 
Stress  is  designated  by  Sx  and  the  two  tangential  component? 


ART.  174 


NORMAL  AND  TANGENTIAL  STRESSES 


451 


Fig.  174 


by  Sxy  and  Sxz.  A  similar  notation  applies  to  each  of  the  other 
faces.  An  S  having  but  one  subscript  denotes  a  tensile  or  com- 
pressive  unit-stress,  and  its 
direction  is  parallel  to  the 
axis  corresponding  to  that 
subscript.  An  S  having  two 
subscripts  denotes  a  shear- 
ing unit-stress,  the  first  sub- 
script designating  the  axis  to 
which  the  face  is  perpendic- 
ular and  the  second  desig- 
nating the  axis  to  which  the 
stress  is  parallel;  thus  Szx  is  on  the  face  perpendicular  to  OZ  and 
its  direction  is  parallel  to  OX.  In  Fig.  174  the  nine  components 
for  three  sides  of  the  parallelepiped  are  shown.  Neglecting  the 
weight  of  the  parallelepiped  the  components  upon  the  three 
opposite  sides  must  be  of  equal  intensity  in  order  that  equilibrium 
may  obtain. 

An  elementary  parallelepiped  in  the  interior  of  a  body  is 
thus  held  in  equilibrium  under  the  action  of  six  normal  and 
twelve  tangential  stresses  acting  upon  its  faces.  The  normal 
stresses  upon  any  two  opposite  faces  must  be  equal  in  intensity 
and  opposite  in  direction.  The  tangential  stresses  upon  any 
two  opposite  faces  must  also  be  equal  in  intensity  and  opposite 
in  direction. 

A  certain  relation  must  also  exist  between  the  six  shearing 
stresses  shown  in  Fig.  174  in  order  that  equilibrium  may  obtain. 
Let  the  parallelopiped  be  a  cube  with  each  edge  equal  to  unity; 
then  if  no  tendency  to  rotation  exists  with  respect  to  an  axis 
through  the  center  of  the  cube  and  parallel  to  OX  it  is  necessary 
that  Syz  should  equal  Sgy.  A  similar  condition  obtains  for  each 
of  the  other  rectangular  axes,  and  hence, 

SXy~Syx  Syz  —  Szy  Sxz=iSzx  (1*4) 

that  is,  those  shearing  unit-stresses  are  equal  which  are  upon 
any  two  adjacent  faces  and  normal  to  their  common  edge. 

The  apparent  unit-stresses  designated  by  5  are  computed  by 


452  MATHEMATICAL  THEORY  OF  ELASTICITY    CHAP,  xix 

the  methods  of  the  preceding  chapters;  it  is  rare,  however,  that 
more  than  three  or  four  of  them  exist,  even  under  the  action  of 
complex  forces.  The  general  problem  is  then  to  find  a  parallele- 
piped such  that  the  resultant  stresses  upon  it  are  wholly  normal. 
These  resultant  normal  stresses  will  be  Si,  £2,  $3,  from  which 
by  (139)  the  true  normal  stresses  TI,  T2,  T3  can  be  found.  It 
will  later  be  shown  that  these  stresses  Si,  £2,  S%  are  the  maxi- 
mum apparent  stresses  of  tension  or  compression  resulting  from 
the  given  normal  and  tangential  stresses. 

Prob.  174.  Let  a,  b,  c  be  the  angles  which  a  line  makes  with  the 
axes  OX,  O  Y,  OZ,  respectively.  Show  that  the  sum  of  the  squares  of 
the  cosines  of  these  angles  is  equal  to  unity. 


ART.  175.    RESULTANT  STRESSES 

The  resultant  unit-stress  upon  any  face  of  the  parallelepiped 
in  Fig.  174  is  the  resultant  of  the  three  rectangular  unit-stresses 
acting  upon  that  face.  Thus  for  the  face  normal  to  the  axis 
OZ  the  resultant  unit-stress  is  given  by, 


and  the  total  resultant  stress  upon  that  face  of  the  parallelepiped 
is  the  product  of  its  area  and  R$. 

The  resultant  unit-stress  R  upon  any  elementary  plane  having 
any  position  can  be  determined  when  the  normal  and  tangential 
stresses  in  the  directions  parallel  to  the  coordinate  axes  are  known. 
Let  a  plane  be  passed  through  the  corners  i,  2,  3,  of  the  paral- 
lelepiped in  Fig.  174,  and  let  a,  b,  c  be  the  angles  that  its  normal 
makes  with  the  axes  OX,  OY,  OZ,  respectively.  Let  a,  /?,  7-, 
be  the  angles  which  the  resultant  unit-stress  R  makes  with  the 
same  axes.  Let  A  be  the  area  of  the  triangle  123;  then  the  total 
resultant  stress  upon  that  area  is  AR,  and  its  components  parallel 
to  the  three  axes  are  AR  cosa,  AR  cos/?,  AR  cos?-.  The  triangle 
whose  area  is  Ay  together  with  the  three  triangles  023,  013,  012, 
form  a  pyramid  which  is  in  equilibrium  under  the  action  of  R 
and  the  stresses  upon  the  three  triangles.  The  areas  of  these 
triangles  are  A  cosa,  Acos£>,  A  cose,  and  the  stresses  upon  them 


ART.  175  RESULTANT  STRESSES  453 

are  the  products  of  the  areas  by  the  several  unit-stresses.     Now 
the  components  of  these  four  stresses  with  respect  to  each  rectan 
;j;ular  axis  must  vanish  as  a  necessary  condition  of  equilibrium. 
Jence,  canceling  out  A,  which  occurs  in  all  terms,  there  results, 

R  cosa  =  Sx   cosa  -f  •  Sxv  cosb  +  Szx  cose 

R  cos/?  =  Sxy  cosa  -f  Sy   cosb  -f  Szy  cose  (1  75) 

R  cosr=Sxz  cosa+Syz  cosb+S2   cose 

in  which  the  second  members  are  all  known  quantities. 

From  these  equations  the  values  of  R  cosa,  R  cos/?,  R  cos  7- 
can  be  computed;  then  the  sum  of  the  squares  of  these  is  R2 
since  cos2a+  cos2/?  +  cos2?-=i.  The  value  of  cosa  is  found  by 
dividing  that  of  R  cosa  by  R,  and  similarly  for  cos/?  and  cosf. 
Now  the  angle  6  between  the  directions  of  R  and  the  normal 
to  the  plane  is  given  by, 

cos/9  =  cosa  cosa+cosfr  cos/?  +  cose 


and  then  the  tensile  or  compressive  unit-stress  normal  to  the 
given  plane  is  R  cos#,  while  the  resultant  shearing  unit-stress 
is  R  sin0.  This  shearing  stress  may  be  resolved  into  two  com- 
ponents in  any  two  directions  on  the  plane. 

As  a  simple  numerical  example,  let  a  bolt  be  subject  to  a 
tension  of  12  ooo  pounds  per  square  inch  and  also  to  a  cross- 
shear  of  8  ooo  pounds  per  square  inch.  It  is  required  to  find 
the  apparent  unit-stresses  on  a  plane  making  an  angle  of  60 
degrees  with  the  axis  of  the  bolt.  Take  OX  parallel  to  the  ten- 
sile force  and  O  Y  parallel  to  the  cross-shear.  Then  Sx  =  +  1  2  ooo, 
Sxy  =  &ooo,  SyX  =  Sooo,  and  the  other  stresses  are  zero;  also 
a  =  3o°,  6  =  60°,  and  £  =  90°.  Then  from  (175), 

R  cosa  =  +  14  390  R  cos/?  =  +6930  R  cosf=o 

and  the  resultant  stress  on  the  given  plane  is, 

R=  (j4  3Qo2+6  9302)*=  15  970  pounds  per  square  inch 
The  direction  made  by  R  with  the  axis  is  now  found: 

cos  a  =1439/1  597  =  0.901  a  =  6^° 

cos/?=  693/1597  =  0.434  /?=25J° 

and  the  angle  included  between  the  resultant  R  and  the  normal 


454  MATHEMATICAL  THEORY  OF  ELASTICITY    CHAP,  xix 

to  the  given  plane  is  computed  by, 

cos#=  0.866X0.901  4-0.5X0.434=0.997 

Lastly,  the  normal  tensile  stress  on  the  plane  is  found  to  be 
R  cos#=i5  920  pounds  per  square  inch,  while  the  shearing  stress 
on  the  plane  is  ^sin#  =  i  200  pounds  per  square  inch. 


Prob.  1  75.  Find  for  the  above  example  the  position  of  a  plane  upon 
which  there  is  no  shearing  stress. 

ART.  176.    THE  ELLIPSOID  OF  STRESS 

Let  the  resultant  unit-stress  R  upon  any  plane  passing  through 
a  given  point  in  the  interior  of  a  stressed  body  make  an  angle 
0  with  the  normal  to  that  plane.  It  will  now  be  shown  that, 
for  different  planes  through  the  given  point,  the  intensity  of  R 
may  be  represented  by  the  radius  vector  of  an  ellipsoid. 

Let  RI,  R2,  RS  be  the  resultant  unit-stresses  upon  the  three 
faces  of  the  parallelepiped  in  Fig.  174,  and  let  61,  #2>  #3  be  the 
angles  which  they  make  with  the  coordinate  axis  OX',  then, 

costfi  =  SX/R!  cos02  =  Syx/R2  cos03  =  SZX/R3 

determine  the  directions  of  RI,  R2,  R%.     Now  let  these  direc- 
tions be  taken  as  those  of  a  new  system  of  oblique  coordinate 
axes,  let  R  be  the  resultant  unit-stress  in  any  direction,  and  let 
Rx,  Ry,  Rz  be  its  components  parallel  to  these  new  axes.    Then 
R  cosa  is  the  component  of  R  parallel  to  OX,  and, 
R  cosa  =  Rx  cos6i  +  Ry  cos02+Rz  cos#3 
or,  inserting  for  the  cosines  their  values,  it  becomes, 


Comparing  this  with  the  first  equation  in  (175),  it  is  seen  that, 

cosa  =  RX/RI  cosb  =  Ry/R2  cose  =  Rz/Rs 

But  the  sum  of  the  squares  of  these  cosines  is  unity;  hence, 


in  which  the  numerators  are  variable  coordinates  and  the  de- 
nominators are  given  quantities.  This  is  the  equation  of  the 
surface  of  an  ellipsoid  with  respect  to  three  coordinate  axes 
having  the  directions  of  RI,  R2,  RZ. 


.  177  THE   THREE   PRINCIPAL   STRESSES  455 

The  surface  of  an  ellipsoid  is  hence  a  figure  whose  radius 
vector  represents  the  resultant  unit-stress  upon  a  plane  the  normal 
to  which  makes  an  angle  0  with  the  direction  ot  that  radius  vector. 
If  the  forces  are  entirely  confined  to  one  plane,  the  ellipsoid  re- 
duces to  an  ellipse. 

If  there  are  three  planes  at  right  angles  to  each  other  which 
are  subject  only  to  normal  stresses,  as  in  Fig.  176,  the  normal 
unit-stresses  Sx,  Sy,  S2  correspond 
to  RI,  7?2,  -^3  in  the  above  equa- 
tion of  the  ellipsoid.  In  this  case 
SXi  S^  Sz  are  the  three  axes  of 
the  ellipsoid.  If  now  shearing 
stresses  are  applied  to  the  faces*  z" 
the  ellipsoid  will  be  deformed,  and 
the  three  axes  will  take  other  posi- 
tions corresponding  to  three  planes  upon  which  no  shearing 
stresses  act.  The  stresses  corresponding  to  the  axes  of  the  ellip- 
soid are  called  principal  stresses. 

Prob.  176.  If  Sy=Sz  in  Fig.  176,  show  that  the  ellipsoid  becomes 
either  a  prolate  spheroid  or  an  oblate  spheroid. 

ART.  177.    THE  THREE  PRINCIPAL  STRESSES 

In  general,  the  resultant  unit-stress  R  upon  a  given  plane 
makes  an  angle  6  with  the  normal  to  that  plane,  and  hence  can 
be  resolved  into  a  normal  stress  of  tension  or  compression  and 
into  two  tangential  shearing  stresses  (Art.  174).  It  is  evident, 
however,  that  planes  may  exist  upon  which  only  normal  stresses 
act,  so  that  6  is  zero  and  R  is  pure  tension  or  compression.  In 
order  to  find  these  planes  and  the  stresses  upon  them,  the  angles 
a,  /?,  r  in  the  equations  (175)  are  to  be  made  equal  to  a,  b,  ct 
respectively.  Also  replacing  R  by  S,  these  equations  become, 

(S— Sx)  cosa  =  Syx  cosb+Sex  cose 
(S-Sy)  cosb=Sxy  cosa+Szy  cose 
(S—Sg)  cosc=Sxz  cosa+SyZ  cosb 

in  which  5,  cosfl,  cos6,  cose  are  four  unknown  quantities.     The 


456  MATHEMATICAL  THEORY  OF  ELASTICITY     CHAP,  xix 

three  angles,  however,  are  connected  by  the  relation, 


and  hence  four  equations  exist  between  four  unknowns. 

Remembering  the  relation  between  the  shearing  stresses  ex- 
pressed in  (174),  the  solution  of  the  equations  leads  to  a  cubic 
equation  for  S,  which  is  of  the  form, 

S3-AS2+BS-C=o  (177) 

in  which  the  values  of  the  three  coefficients  are, 
A  =  Sx  +  Sy  +  Sz, 

Jj  =  o  xo  y  ~T"  O  y^z  i   <^z^  x      *^   xy      ^   yz      *-^   zx 

C  =  0^0^021"  Z^xy^yz^zx      ^x1^   yz      &y&  Zx      ^2^  xy 

and  the  three  roots  of  this  cubic  equation  are  the  three  normal 
stresses  of  tension  or  compression,  which  are  called  the  three 
principal  unit-stresses  and  represented  by  Si,  S2,  S3. 

The  directions  of  these  principal  unit-stresses  Si,  52,  S3,  with 
respect  to  the  rectangular  axes  OX,  OF,  OZ,  are  given  by  the 
values  of  cos  a,  cos  b,  cose,  which  are  found  to  be, 

cosa=(f»j/f»)*  cos&=  (/»2/w)*  cosc=(w3/w)* 

in  which  mi,  m2,  w3,  m,  represent  the  following  functions  of  the 
given  and  principal  unit-stresses: 

mi  =  (Sv  -  S)  (Sz  -  S)  -  S2yz  m2  =  (Sz  -  S)  (Sx  -S)-  S2ZX 


and  it  will  now  be  shown  that  each  principal   stress  is  perpen- 
dicular to  the  plane  of  the  other  two. 

Let  Si,  52,  53  be  the  three  roots  of  the  cubic  equation  (177). 
Let  fli,  &i,  c\  be  the  angles  which  Si  makes  with  the  three  co- 
ordinate axes  OX,  OY,  OZ,  and  let  a2,  b2,  c2  be  the  angles  which 
52  makes  with  the  same  axes.  The  angle  between  the  direc- 
tions of  Si  and  S2  is  then  given  by, 

cosc/;>  =  cosai  cosa2+cosfri  cosb2+cosci  cosc2 

Now  in  the  first  set  of  formulas  of  this  article  let  5  be  made  Si 
and  a,  b,  c  be  changed  to  ai,  biy  GI\  let  the  first  equation  be  mul- 
tiplied by  cos<i2,  the  second  by  cos62,  and  the  third  by 
and  let  the  three  equations  be  added;  then, 


ART.  178  MAXIMUM    SHEARING    STRESSES  457 

is  one  term  in  this  sum.  Again,  let  S  be  made  £2,  an  da,  b,  c, 
be  changed  to  a2,  b2,  c2\  let  the  equations  be  multiplied  by  costfi, 
cos^i,  cosci,  respectively,  and  added;  then, 

S2(cosai  cosaa+cos^i  cos&o+cosci  cos^) 

is  one  term  in  the  sum,  while  all  the  other  terms  are  the  same 
as  before.  Hence  if  Si  and  S2  are  unequal,  the  factor  in  the 
parenthesis,  which  is  cos<£,  must  vanish;  $  is  therefore  a  right 
angle  or  Si  and  S2  are  perpendicular.  In  the  same  manner  it 
may  be  shown  that  $3  is  perpendicular  to  both  Si  and  S2. 

The  three  principal  stresses  are  hence  perpendicular  to  each 
other,  and  as  the  only  diameters  of  the  ellipsoid  which  have 
this  property  are  its  axes,  it  follows  that  the  directions  of  the 
principal  stresses  Si,  £2,  £3  are  those  of  the  axes  of  the  ellip- 
soid of  stress.  These  principal  stresses  thus  give  the  apparent 
maximum  normal  stresses  of  tension  or  compression;  from  (139) 
the  corresponding  true  unit-stresses  T\,  T2,  T%  are  then  found. 

An  interesting  property  of  the  three  rectangular  stresses  Sx, 
Sy,  Sg,  is  that  their  sum  is  constant,  whatever  may  be  the  posi- 
tion of  the  coordinate  axes.  For,  the  sum  of  the  three  princi- 
pal stresses  Si,  S2,  Sa  is  equal  to  the  coefficient  A  in  the  cubic 
equation  of  (177),  and  hence, 


that  is,  the  sum  of  the  normal  unit-stresses  in  any  three  rectan- 
gular directions  is  constant. 

Prob.  177.  When  two  principal  stresses  are  equal,  show  that  the 
value  of  each  is  (AB-9C)/(2A2-6B),  where  A}  B,  C  are  the  coeffi- 
cients in  (177). 

ART.  178.    MAXIMUM  SHEARING  STRESSES 

As  there  are  certain  planes  upon  which  the  tensile  and  com- 
pressive  unit-stresses  are  a  maximum,  so  there  are  certain  other 
planes  upon  which  the  shearing  unit-stresses  have  their  maxi- 
mum values.  In  order  to  determine  these  it  is  well  to  take  the 
axes  of  the  ellipsoid  as  the  coordinate  axes,  and  upon  the  planes 
normal  to  these  there  are  no  shearing  stresses.  The  stresses 


458  MATHEMATICAL  THEORY  OF  ELASTICITY    CHAP.  XIX 

Si,  S2,  S3  will  give  apparent  shearing  stresses  on  other  planes, 
while  TI,  T2.  T3  will  give  the  true  shearing  stresses. 

Let  i  2  3  in  Fig.  178  be  any  plane  whose  normal  makes  the 
angles  a,  b,  c  with  the  coordinate  axes.  Let  R  be  the  resultant 
unit-stress  upon  this  plane,  and  a,  ft,  7-  be  the  angles  which 
it  makes  with  the  same  axes.  The  angle  between  R  and 
the  normal  to  the  plane  is  expressed  by, 

cos#=cosa  cosa  +  cosft  cos/?+cosc 


and    the    resultant    shearing    unit- 
stress  on  the  plane  is, 


If   R   be   apparent  stress,   this    is 
the    apparent    shearing    stress;    if 
R    be    true  stress,  this  is  the  true 
shearing  stress  which  acts  along  the  plane. 

The  value  of  R,  as  a  true  stress,  is  given  by, 

R2=(Ti  cosa)2+(T2  cosb)2+(T3  cose)2 

Now,  since  both  R  cosa,  and  TI  cosa  are  components  of  R  in 
the  direction  OX,  they  are  equal,  and  hence, 

cosa=  (Ti/R)  cosa  cos/?=  (T2/R)  cosb          cosj-=  (T3/R)  cose 

Substituting  these  in  the  value  of  cos#,  the  resulting  true  shear- 
ing unit-stress  is  expressed  by, 

J2=  (71!  cosa)2  +  (T2  cosb)2  +  (T3  cosc)2-(Ti  cos2a  +  T2  cos2b  +  T3  cosV)2 

by  the  discussion  of  which  the  values  of  a,  b,  c,  which  render  T 
a  maximum,  are  deduced.  Bearing  in  mind  that  the  sum  of 
the  squares  of  the  three  cosines  is  unity,  the  discussion  gives, 

c=9o°  a=b=±4$°  T=±$(Ti-T2) 

and  therefore  there  are  six  planes  of  maximum  shearing  stress, 
each  of  which  is  parallel  to  one  of  the  principal  stresses  and 
bisects  the  angle  between  the  other  two.  On  each  of  these  planes 


ART.  179  DISCUSSION   OF   A   CRANK   PlN  459 

the  shearing  unit-stress  is  one  half  the  difference  of  the  principal 
unit-stresses  whose  directions  are  bisected. 

The  same  investigation  applies  equally  well  to  the  apparent 
shearing  unit-stresses,  whose  maximum  values  are, 


and  whose  directions  are  the  same  as  those  of  the  maximum 
true  shearing  unit-stresses.  The  .sign  ±  indicates  that  the  shears 
have  opposite  directions  on  opposite  sides  of  the  plane,  but  in 
numerical  work  it  is  always  convenient  to  take  them  as  positive 
or,  rather,  as  signless  quantities. 

As  an  example,  let  a  bar  be  subject  to  a  tension  of  3  ooo 
pounds  per  square  inch  in  the  direction  of  its  length  and  to  a 
compression  of  6  ooo  pounds  per  square  inch  upon  two  oppo- 
site sides.  Here  Si  =+3  ooo,  £2=—  6000,  53=0;  then  the 
maximum  apparent  shearing  stresses  are  4  500,  3  ooo,  and  i  500 
pounds  per  square  inch.  But  from  (139),  taking  X  as  -J,  the  true 
tensile  and  compressive  stresses  as  T\  =  +  5  ooo,  T^  =  —  7  ooo, 
r3=+i  ooo  pounds  per  square  inch,  and  then  from  (178)  the 
maximum  true  shearing  stresses  are  6  ooo,  4  ooo,  and  2  ooo 
pounds  per  square  inch. 

Prob.  178.  Compute  the  maximum  apparent  and  true  shearing  unit- 
stresses  for  a  cast-iron  parallelepiped,  2X4X8  inches,  which  is  subject 
to  compression  of  3200  pounds  upon  its  largest  faces,  60  pounds  upon 
its  smallest  faces,  and  500  pounds  upon  the  other  faces. 

ART.  179.     DISCUSSION  OF  A  CRANK  PIN 

To  apply  the  preceding  principles  to  a  particular  case,  a 
crank  pin  similar  to  that  investigated  in  Art.  98  may  be  taken. 
The  axis  OX  is  assumed  parallel  to  the  axis  of  the  pin,  O  Y  parallel 
to  the  crank  arm,  and  OZ  perpendicular  to  both,  the  notation 
being  the  same  as  in  Fig.  174.  On  one  side  of  the  crank  pin 
near  its  junction  with  the  arm  there  were  found  the  following 
apparent  stresses:  A  cross-shear  from  the  pressure  of  the  con- 
necting rod  giving  £^  =  300  pounds  per  square  inch,  a  shear 
due  to  the  transmitted  torsion  giving  Sxz  =  <^oo  pounds  per  square 


460  MATHEMATICAL  THEORY  OF  ELASTICITY    CHAP.  XIX 

inch,  a  flexural  stress  due  to  the  connecting-rod  giving  Sx=  +800 
pounds  per  square  inch,  a  flexural  stress  due  to  the  transmitted 
torsion  giving  Sx  =  + 1  600  pounds  per  square  inch,  and  two 
compressions  due  to  shrinkage  giving  Sy=—  4000  and  Sz  = 
—  2  ooo  pounds  per  square  inch. 

The  two  shears  having  the  same  direction  add  together,  as 
also  the  two  tensions,  and  the  data  then  are, 

Sxz=  1 200          Sx  =  +  2400          Sy  =  —  4000          Sz=—  2000 
Inserting  these  in  the  cubic  equation  (177)  it  becomes, 

•S3+3  6oo,S2—  7  840  oooS—  27  200  ooo  000=0 

and  its  three  roots  are  the  three  principal  stresses.  To  solve 
this  equation,  put  S  =  x  —  i  200,  and  it  reduces  to, 

xP— 12  1 60  ooox—ii  096  ooo  000=0 

As  this  cubic  equation  has  three  real  roots,  it  is  to  be  solved  by 
the  help  of  a  table  of  cosines ;  thus  let, 

y2=  12  1 60  ooo  2r3  cos3^>=  ii  096  ooo  ooo 

from  which  ^  =  2013  and  cos3<£  =0.6801.  Then  from  Table  17 
is  found  3<£  =  47°  09',  whence  ^  =  15°  43'.  The  roots  are  now 
computed  as  below,  and  by  subtracting  i  200  from  each  the  three 
principal  stresses  are  ascertained: 

xi  =  2rcos(j)  =+3880  Si  =+2680 

x2=2r  cos(<5&+ 120°)  =  —2890  $2=—  4090 

x%=2r  005(^  +  240°)  =—  990  53=— 2190 

and  these  are  the  apparent  principal  stresses  in  pounds  per  square 
inch.  Taking  e  =  J  for  steel,  the  true  principal  stresses  are  now 
found  by  (139)  to  be, 

^1=4-4770  r2=-425o  r3=-i72o 

which  show  that  the  maximum  true  tensile  stress  is  nearly  double 
the  apparent,  while  the  maximum  true  compressive  stress  is 
6  percent  greater  than  the  apparent. 

An  ordinary  solution  of  this  problem,  in  which  no  combina- 
tion of  stresses  was  made,  would  show  the  greatest  tension  to 
be  2  400  pounds  per  square  inch,  while  the  complete  solution  as 
above  given  shows  that  the  greatest  true  tension  is  nearly  twice 
as  great.  The  ordinary  solution  shows  the  shearing  stress  to 


ART.  180 


THE  ELLIPSE  OF  STRESS 


461 


be  i  200  pounds  per  square  inch,  but  by  applying  (178)  to  the 
above  values  of  T  it  is  seen  that  the  maximum  true  shearing  stress 
is  4510  pounds  per  square  inch.  It  thus  appears  that  where 
many  stresses  combine,  as  at  the  junction  of  a  crank  pin  with 
its  web,  the  common  methods  of  investigation  give  unit-stresses 
which  are  far  too  small;  it  also  follows  that  designs  for  such  cases, 
made  by  ucing  the  common  methods,  should  be  based  upon 
low  unit-stresses. 

Prob.  179.  Apply  the  cubic  equation  (177)  to  the  case  of  a  bar  acted 
upon  only  by  the  tensile  unit-stress  Sx  and  the  transverse  shearing  unit- 
stress  Sxy.  Deduce  the  principal  stresses  for  this  case. 

ART.  180.     THE  ELLIPSE  OF  STRESS 

The  ellipse  of  stress  is  that  particular  case  where  one  of  the 
principal  stresses  is  zero,  in  which  event  the  last  term  of  (177) 
vanishes.  An  instance  of  this  is  where  S2  =  o,  Syz  =  o,  Sxz  =  o, 
which  is  that  of  a  body  subject  to  the  normal  unit-stresses  5*, 
Syj  and  to  the  shearing  unit-stress  Sxy.  The  cubic  equation 
then  reduces  to  the  simple  quadratic, 

,S2~  (Sx  +  Sy)S+SxSy~S2xy=0 

and  the  two  roots  of  this  are  the  two  principal  apparent  stresses 
whose  directions  correspond  to  the  two  axes  of  the  ellipse.  From 
this  quadratic  equation,  the  formulas  (144)  were  derived. 


B 


Fig.  180 

Let  Si  and  S2  be  these  roots,  and  in  Fig.  180  let  OA  and  OB 
be  laid  off  at  right  angles  to  represent  their  values.  Let  an 
ellipse  be  described  upon  the  axes  AA  and  BB,  and  let  $  be 
the  angle  AON  which  any  line  ON  makes  with  OA.  Upon  a 


462  MATHEMATICAL  THEORY  OF  ELASTICITY    CHAP.  XIX 

plane  normal  to  ON  at  O  the  normal  unit-stress  of  tension  or 
compression  has  the  value, 


and  the  tangential  shearing  unit-stress  is, 

OS=(Si-Sn)  sin<£  cos<£ 

while  the  resultant  of  these  gives  the  resultant  unit-stress, 
OR=(Si2  cos2<£  +  S22  sin2^)* 

The  diagrams  in  Figs.  180  give  graphic  representations  of  these 
values  as  the  angle  (j>  varies  from  c  to  360  degrees.  In  the  first 
diagram  Si  and  £2  are  both  tension  or  both  compression,  in  the 
second  diagram  one  is  tension  and  the  other  compression.  The 
broken  curve  shows  the  locus  of  the  point  AT,  and  the  dotted 
curve  the  locus  of  S.  For  every  value  of  <£  the  lines  OS  and  ON 
are  at  right  angles  to  each  other,  and  OR  is  their  resultant. 

As  a  simple  example,  take  the  case  of  a  bolt  subject  to  an 
axial  tension  of  2  ooo  and  also  to  a  cross-shear  of  3  ooo  pounds 
per  square  inch.  Here  Sx=+2Ooo,  Sxy  =  ^ooo)  and  Sy  =  o; 
the  above  quadratic  equation  then  gives  Si  =+4160  and  52  = 

—  2  160  pounds   per  square  inch  for  the  two  maximum   unit- 
stresses  of  tension  and  compression.     The  direction  made  by  Si 
with  the  axis  of  the  bolt,  as  found  by  the  value  of  cosa  in  Art.  175, 
is  about  54  J  degrees.     From  (178)'  the  maximum  shear  is  3  160 
pounds   per  square   inch.     These   are   the   apparent  maximum 
unit-stresses. 

To  find  the  true  maximum  stresses,  formulas  (139)  give, 
taking  J  as  the  factor  of  lateral  contraction,  T\  =  +  4  880,  T%  = 

—  3550,    TS=  —  670   pounds   per  square  inch  as   the  principal 
tensions  and  compressions;  then  from  (178)  the  greatest  shearing 
stress  is  T  =  4  220  pounds  per  square  inch.     Here  the  true  maxi- 
mum tension  is  17  percent  greater  than  the  apparent,  the  true 
compression  is  64  percent  greater,  and  the  true  shearing  stress 
is  33  percent  greater.     The  true  stresses  cannot  be  represented 
by  an  ellipse,  but  an  ellipsoid  of  internal  stress  results  of  which 
the  second  diagram  in  Fig.  180  may  be  regarded  as  a  typical 
section. 


ART.  181  SHEARING  MODULUS  OF  ELASTICITY  463 

Cases  can,  however,  be  imagined  in  which  one  of  the  true 
principal  stresses  is  zero.  If  Si,  £2,  £3  are  the  apparent  stresses 
in  three  rectangular  directions,  it  is  seen  from  (139)  that  when 
S3  —  eSi  —  £52  is  zero,  the  true  stress  7^  is  also  zero.  For  instance, 
let  a  cube  be  under  compression  by  three  normal  stresses  of  30, 
24,  and  18  pounds  per  square  inch  and  let  £  =  J;  then  7\  =  i6, 
T2  =  8,  and  7^=0.  Here  the  ellipse  of  true  stress  has  its  correct 
application  and  there  are  no  true  stresses  in  a  plane  normal  to 
the  plane  of  T\  and  TV 

Prob.  180.  A  body  is  subject  to  a  tension  of  4000  and  to  a  com- 
pression of  6  ooo  pounds  per  square  inch,  these  acting  at  right  angles 
to  each  other.  Construct  the  ellipse  of  apparent  stresses  and  find  the 
positions  of  two  planes  on  which  there  are  no  tensile  or  compressive 
stresses. 

ART.  181.     SHEARING  MODULUS  OF  ELASTICITY 

The  shearing  modulus   of  elasticity  F,   denned   in  Art.  15, 
must  have  a  relation  to  the  modulus  E  for  tension  or  compression, 
since  the  action  of  shear  upon  a  body  produces  internal  tensile 
and  compressive  stresses  (Art.  143).     Let  Fig.  181  represent  the 
face  of  a  cube  which  is  acted  upon  by  a 
vertical  shear  S,  the  edge  of  the  cube  being 
unity   so   that   the   vertical  shearing   unit- 
stress  is  also  S.     Under  the  action  of  this 
shear,   the  face  of  the  cube  becomes  dis- 
torted, as  shown  greatly  exaggerated  by  the 
broken  lines,   and  the  longer  diagonal  of 
the  rhombus  is  under  a  tensile  unit-stress 
while  the  shorter  one  is  under  a  compressive  unit-stress,  each 
of  these  being  equal  to  S,  as  proved  in  Art.  143.     Let  s  be  the 
distortion  parallel  to  the  shear  S;    then  e  =  S/F  from  the  defini- 
tion of  shearing  modulus  of  elasticity. 

The  longer  diagonal  of  the  square  has  the  length  2*  and 
after  the  distortion  its  length  becomes  [i  +  (i  +  e)2?;  by  using 
the  approximate  method  for  extracting  roots  explained  in  Art.  13, 
and  neglecting  the  square  of  e,  this  reduces  to  2*(i-f-J$).  The 


464  MATHEMATICAL  THEORY  OF  ELASTICITY     CHAP.  XIX 

change  in  length  of  this  diagonal  hence  is  2*  .  Je,  and  the  unit- 
elongation  is  Je.  In  a  similar  manner  the  length  of  the  shorter 
diagonal  after  the  distortion  is  2*(i  —  Je)  and  the  unit-shortening 
is  Je.  Accordingly  the  change  per  unit  of  length  for  each  diag- 
onal is  \S/F. 

Now  let  A  be  the  factor  of  lateral  contraction  (Art.  13)  the 
mean  value  of  which  is  J  for  cast  iron  and  J  for  wrought  iron 
and  steel.  When  a  body  is  acted  upon  by  a  tension  producing 
the  unit-stress  S,  there  results  a  unit-elongation  S/E  and  a  lateral 
unit-shortening  XS/E.  When  a  body  is  acted  upon  a  tension 
producing  the  unit-stress  S  and  by  a  compression  at  right  angles 
producing  the  same  unit-stress  S,  the  unit- elongation  is  (i  +X)S/E, 
as  is  shown  in  Art.  139;  the  lateral  unit-shortening  has  also  the 
same  value.  Accordingly,  for  the  case  of  Fig.  181,  each  diagonal 
has  suffered  a  change  per  unit  of  length  equal  to  (i+X)S/E. 

The  change  per  unit  of  length  for  each  of  the  diagonals  of  the 
face  of  the  cube  has  now  been  found  by  two  different  methods; 
equating  the  two  values,  there  results, 

E=2(i  +  X)F  or  F=E/2(i  +  X)  (181) 

which  give  the  relation  between  the  two  moduluses.  Hence  when 
E  and  A  have  been  determined  by  measurements  on  a  bar  under 
tension,  the  shearing  modulus  of  elasticity  may  be  computed. 

Using  the  mean  values  of  E  given  in  Art.  9,  and  the  mean  values 
of  A  as  above  stated,  the  mean  values  of  the  shearing  modulus 
of  elasticity  are  found  to  be  as  follows  for  iron  and  steel: 

for  cast  iron,  F—   6  ooo  ooo  pounds  per  square  inch 

for  wrought  iron,       F=  9  400  ooo  pounds  per  square  inch 
for  steel,  F=n  200  ooo  pounds  per  square  inch 

and  these  have  been  verified  by  experiments  on  the  torsion  of 
shafts.  There  is  little  known  regarding  the  values  of  A  for  other 
materials,  and  it  may  be  said  that  formula  (181)  does  not  apply 
to  fibrous  or  non-homogeneous  materials  for  the  reason  that  E 
is  not  the  same  in  different  directions.  It  is  not  to  be  expected 
then  that  F  could  be  correctly  computed  for  timber  from  a  value 
of  E  obtained  from  a  tension  parallel  to  the  grain. 


ART.  182  THE  VOLUMETRIC  MODULUS  465 

Four  different  methods  are  available  for  determining  the  shear- 
ing modulus  of  elasticity;  first,  by  the  measurement  of  the  detru- 
sion  per  unit  of  length  in  a  short  bar  or  bearh  like  Fig.  181 ;  second, 
from  the  angle  of  twist  of  a  shaft  (Art.  93) ;  third,  from  the  deflec- 
tions of  beams  of  different  lengths  and  sizes  (Art.  125) ;  and 
fourth,  by  the  use  of  formula  (181).  When  sources  of  error 
are  eliminated  from  the  experiments,  these  different  methods 
give  results  for  F  which  agree  very  well  for  homogeneous  mate- 
rials. A  fifth  method;  which  is  really  an  extension  of  the  fourth, 
will  be  explained  in  the  next  article.  All  these  methods,  of 
course,  apply  only  when  the  shearing  elastic  limit  of  the  material 
is  not  exceeded  by  the  unit-stress. 

Prob.  181.  A  bar  of  steel,  0.5050  inches  in  diameter  and  2.0000 
long,  is  observed  to  be  2.0013  inches  long  when  under  a  tension  of 
4  ooo  pounds,  while  its  diameter  is  then  found  to  be  0.5047  inches. 
Compute  the  shearing  modulus  of  elasticity. 

ART.  182.    THE  VOLUMETRIC  MODULUS 

The  modulus  of  elasticity  E  for  axial  tension  or  compression 
is  defined  to  be  the  ratio  of  the  longitudinal  unit-stress  S  to  the 
unit-elongation  or  unit-shortening  e;  thus  E  =  S/s.  Similarly, 
the  volumetric  modulus  of  elasticity,  which  will  be  represented 
by  G,  is  the  ratio  of  a  unit-stress  S  which  acts  in  all  directions 
upon  the  body  to  the  change  per  unit  of  volume.  Thus,  if  a 
uniform  unit-pressure  acts  upon  a  body  of  volume  unity  and 
produces  the  change  e'  in  that  volume,  then  G  =  S/£f.  It  is 
required  to  find  the  values  of  e'  and  G,  and  also  the  relation 
between  G  and  E. 

Let  A  be  the  factor  of  lateral  contraction  of  the  homogeneous 
cube,  each  edge  of  which  is  unity,  while  each  face  is  subject  to 
the  same  pressure  S.  Then,  from  Art.  139,  the  unit-shortening 
of  each  edge  of  the  cube  is  £'  =  (S-2>IS)/E;  or  since  e  represents 
S/E,  the  unit-shortening  is  fi'  =  (i  — 2A)e.  The  volume  of  the 
cube,  which  was  originally  unity,  now  becomes  (i  — e')3,  and 
hence  the  change  in  volume  is  35'  when  e'  is  so  small  that  its 
square  and  cube  may  be  neglected.  The  change  per  unit  of  volume 
is  then  three  times  the  change  per  unit  of  length  of  each  edge  of 


466  MATHEMATICAL  THEORY  OF  ELASTICITY    CHAP,  xix 

the  cube;  hence  e'  =  3(1-2^)5  gives  the  change  per  unit  of 
volume,  and  hence  the  volumetric  modulus  G  is  5/3(1—  2^)e, 
in  which  £  is  the  change  per  linear  unit  due  to  an  axial  unit- 
stress  S  of  tension  or  compression  only;  accordingly, 

G-S/e'  G=S/3(i-2X)e  G=E/$(i-2X)       (182) 

are  formulas  for  G,  of  which  the  third  gives  the  relation  between  G 
and  E.  For  example,  A  is  about  J  for  steel,  and  hence  the  volumetric 
modulus  for  steel  is  equal  to  the  modulus  for  tension  or  compression. 

In  the  last  article  the  relation  between  the  shearing  modulus 
F  and  the  tensile  or  compressive  modulus  E  was  deduced.  The 
last  formula  of  (181)  and  the  last  formula  of  (182)  furnish  two 
equations  from  which,  by  the  elimination  of  ^,  there  is  found 
as  the  relation  between  £,  F,  G;  hence, 

F=  3£G/(9G-E)  G=  EF/(9F-  3£) 

give  the  value  of  each  modulus  in  terms  of  the  other  two.  For 
example,  let  it  be  known  that  for  cast  iron  £  =  15,000000  and 
F  =  6  ooo  ooo  pounds  per  square  inch ;  then  G  =  10  ooo  ooo  pounds 
per  square  inch.  The  last  formula  shows  that  F  cannot  be  as 
small  as  %E,  for  G  becomes  infinite  when  E  is  equal  to  ^F. 

Water  is  matter  which  propagates  stress  in  all  directions  so 
that  a  unit-pressure  S  applied  to  the  surface  of  a  column  of  water 
produces  a  resisting  unit-pressure  S  on  all  the  confining  surfaces. 
According  to  the  experiments  made  by  Grassi  in  1850,  the  de- 
crease in  a  unit-volume  of  water  caused  by  the  pressure  of  one 
atmosphere,  or  14.7  pounds  per  square  inch,  has  a  mean  value 
of  0.00005;  hence  the  mean  volumetric  modulus  of  elasticity 
for  water  is  G  =  14. 7/0.00005  =  294  ooo  pounds  per  square  inch, 
which  is  about  one  one-hundredth  of  that  of  steel,  so  that  water  is 
about  loo  times  more  compressible  than  is  steel  within  its  elastic 
limit.  Water  has  no  proper  value  of  E,  because  it  is  impossible 
for  a  column  to  be  subjected  to  longitudinal  pressure  only;  when 
water  in  a  pipe  is  under  axial  pressure,  the  shortening  that  is 
measured  is  due  to  a  unit-pressure  acting  laterally  as  well  as 
axialiy,  and  this  gives  the  decrease  in  volume  if  the  walls  of  the 
pipe  are  unyielding. 


AJIT.  183  STORED  INTERNAL  ENERGY  467 

Prob.  182.  Prove,  for  a  homogeneous  solid,  that  the  ratio  G/F  is 
equal  to  2(1  +  ^)73(1  —  2^).  Also  that  the  value  of  X  may  be  expressed 
by  J-E/6G. 

ART.  183.     STORED  INTERNAL  ENERGY 

The  cases  of  resilience,  or  stored  internal  energy,  which  were 
discussed  in  Chapter  XIII,  relate  only  to  simple  axial  stress  and 
to  simple  shear;  when  a  body  is  subject  to  several  external 
forces  acting  in  different  directions,  the  expressions  for  resilience 
become  more  complex.  Fig.  183  represents  a  parallelepiped, 
the  faces  of  which  are  acted  upon  only 
by  the  normal  unit-stresses  Si,  £2,  83. 
Let  this  parallelepiped  be  homogeneous 
so  that  the  modulus  of  elasticity  E  and 
the  factor  of  lateral  contraction  A  are 
the  same  in  all  directions.  Let  l\  be  the 
length  parallel  to  Si,  and  a\  the  section 
area  normal  to  Si;  then  the  total  stress  on  this  section  area  is 
0iSi,  and  from  (10)  and  (139)  the  change  in  the  length  of  /i 
is  /i(5i-~A52-AS3)/E.  The  work  done  while  the  stress  a^Si 
is  increasing  from  zero  up  to  its  final  value  is  one-half  the  product 
of  the  stress  and  change  of  length,  provided  the  elastic  limit 
of  the  material  is  not  exceeded  (Art.  119);  hence  the  stored  in- 
ternal energy  due  to  Si  is  %all1(S]2-XSiS2-lSiS3)/E  and  this  is 
proportional  to  the  volume  aji.  A  similar  expression  may  be 
written  for  the  energy  due  to  $2  and  another  for  that  due  to  Sa; 
and  the  sum  of  the  three  is, 


in  which  V  denotes  the  volume  of  the  parallelepiped.     Here  the 
sign  of  each  5  is  positive  for  tension  and  negative  for  compression. 

A  discussion  of  this  formula  shows  that  K  has  usually  a 
smaller  value  when  the  signs  of  Si,  S2,  S3  are  the  same  than  when 
one  has  a  sign  opposite  to  that  of  the  other  two.  When  the  unit- 
stresses  are  equal  in  sign  and  magnitude,  then  K—^(i  —  2A)V.  S2/E  ; 
for  steel  X  is  J  and  the  resilience  becomes  K  =  %(S2/E)V,  which 
is  the  same  as  for  simple  axial  stress  (Art.  120).  As  a  numerical 


468  MATHEMATICAL  THEORY  OF  ELASTICITY     CHAP.  XIX 

example,  let  the  three  edges  of  the  parallelepiped  01,  02,  03  be 
8,  6,  and  4  inches  long  and  the  material  be  cast  iron  for  which 
A  is  J,  and  let  the  unit-stresses  Si,  S2,  Ss  be  3  ooo,  4  ooo?  5  ooo 
pounds  per  square  inch  compression;  then  K  =  I>JO  inch-pounds 
=  14  foot-pounds  is  the  stored  internal  energy. 

When  the  above  parallelopiped  is  subject  to  the  action  of 
six  pairs  of  shears  of  which  the  unit-stresses  are  Si2,  £23,  Ssi, 
in  the  notation  of  Art.  174,  the  corresponding  unit-detrusions 
are  %S2i2/F,  %S22s/F,  %S23i/F,  where  F  is  the  shearing  modulus 
of  elasticity,  and  the  sum  of  these  multiplied  by  the  volume  of 
the  body  gives, 

^/=JF(S2i2+S223  +  S23l)/F  (183)' 

as  the  stored  energy  due  to  the  shears.  This  is  to  be  added  to  the 
value  of  K  in  (183)  when  normal  stresses  also  act  upon  the  faces 
of  the  parallelopiped.  A  comparison  of  the  two  formulas  shows 
that,  if  a  body  is  acted  upon  by  normal  and  tangential  forces  of 
equal  intensity  in  three  rectangular  directions,  the  stored  in- 
ternal energy  due  to  shearing  may  often  be  greater  than  that 
due  to  the  normal  stresses. 

The  principle  of  least  work,  established  in  Art.  126,  states 
that  the  internal  stresses  which  prevail  in  a  body  under  the  action 
of  external  forces  are  those  which  render  the  stored  internal  energy 
a  minimum.  This  principle  may  be  used  in  connection  with 
(183)  to  determine  the  stresses  Si,  S2,  $3  in  cases  where  the 
conditions  of  static  equilibrium  are  insufficient  in  number.  For 
example,  it  was  assumed  in  Art.  163  that  the  tangential  and  radial 
stresses  in  a  spherical  annulus  were  connected  by  the  relation 
2S  +  R  =  3i  constant.  This  assumption  may  seem  an  arbitrary 
one,  but  it  can  be  shown  by  an  algebraic  investigation,  which 
is  too  lengthy  to  be  given  here,  that  the  total  stored  internal 
energy  is  less  when  28  +  R  is  constant  throughout  the  spherical 
anuulus  than  when  this  sum  varies  according  to  any  function 
of  the  distance  x  from  the  center. 

The  ether  of  space  transmits  light,  electricity,  and  gravitation 
from  one  body  to  another.  The  phenomena  of  gravitation  are 
familiar  to  every  one,  but  the  explanation  of  its  cause  has  not  yet 


ART.  183  STORED    INTERNAL   ENERGY  469 

been  discovered.  All  observations  and  theory  indicate  that  the 
ether  is  an  elastic  substance  which  obeys  the  laws  of  the  mathe- 
matical theory  of  elasticity.  Accordingly  it  seems  that  the 
general  conclusions  of  Art.  163  regarding  the  distribution  of 
stresses  in  hollow  spheres  should  apply  to  those  stresses  in  the 
ether  which  cause  the  mutual  gravitation  of  bodies  of  matter; 
if  this  be  so,  these  stresses  vary  inversely  as  the  cube  of  the  dis- 
tance. The  actual  forces  of  gravitation,  however,  vary  inversely 
as  the  square  of  the  distance,  and  it  is  not  easy  to  see  how  this 
law  is  deduced  from  that  of  the  distribution  of  the  stresses.  To 
solve  the  great  riddle  of  gravitation,  a  more  definite  knowledge 
is  required  regarding  the  constitution  of  matter,  and  the  indi- 
cations are  that  an  explanation  may  be  obtained  during  the 
twentieth  century. 

Prob.  183.  Consult  Isenkrahe's  Das  Rathsel  von  der  Schwerkraft 
(Braunschweig,  1879)  for  critical  reviews  of  the  various  attempts  to 
explain  the  phenomena  of  gravitation. 


470  TESTING  OF  MATERIALS  CHAP.  XX 


CHAPTER  XX 

TESTING  OF   MATERIALS 
ART.  184.    TESTING  MACHINES 

The  first  experiments  on  the  strength  of  materials  were  made 
on  the  rupture  of  beams  of  timber.  A  picture  in  Galileo's  Discorsi 
(Leiden,  1638),  shows  a  cantilever  beam  projecting  from  a  wall 
and  loaded  with  a  weight  at  the  free  end,  and  it  was  probably 
from  experiments  of  this  kind  that  Galileo  was  led  to  the  con- 
clusion that  the  strength  of  rectangular  beams  varies  as  the 
squares  of  their  depths.  During  the  eighteenth  century  experi- 
ments were  made  in  France  on  timber  in  flexure  and  tension, 
only  questions  of  ultimate  strength  being  considered,  while  the 
elastic  limit  was  unrecognized.  Hooke's  experiments  on  springs, 
from  which  he  deduced  the  law  of  proportionality  between  stress 
and  elongation  had,  indeed,  been  announced  in  1678,  but  it  was 
not  until  1798  that  Girard  made  the  first  comprehensive  series 
of  experiments  on  the  elastic  properties  of  beams.  Nearly  a 
quarter  of  a  century  later  Barlow,  Tredgold  and  Hodgkinson 
experimented  on  timber  and  cast  iron  both  in  the  form  of  beams 
and  columns;  their  methods  and  results,  although  now  seem- 
ingly rude  and  defective,  are  deserving  of  praise  as  the  first  of 
real  practical  value. 

In  1849  was  published  in  London  the  '  Report  of  the  Commis- 
sioners on  the  Application  of  Iron  to  Railway  Structures,'  which 
may  be  regarded  as  the  landmark  of  the  beginning  of  the  modern 
methods  of  testing.  The  immediate  result  of  this  report  was  the 
decision  of  the  English  board  of  trade  that  the  factor  of  safety 
for  cast  iron  should  be  twice  as  great  for  rolling  loads  as  for  steady 
ones,  while  throughout  Europe  and  the  United  States  it  aroused 
marked  impetus  in  the  subject  of  testing  materials. 


ART.  184  TESTING  MACHINES  471 

The  first  testing  machines  in  the  United  States  were  those 
built  by  Wade  and  Rodman  between  1850  and  1860  for  testing 
gun  metal.  About  this  time  the  rapid  introduction  of  iron  bridges 
led  to  experiments  by  Plympton  and  by  Roebling.  Prior  to  1865 
apparatus  was  built  by  each  experimenter  for  his  special  work, 
but  in  that  year  Fairbank  put  upon  the  market  the  first  testing 
machines  for  commercial  work.  A  little  later  the  machines  of 
Olsen  and  of  Riehle  for  tensile,  compressive,  and  flexural  tests 
were  introduced  and  have  since  been  widely  used.  The  machine 
devised  by  Emery,  soon  after  1875,  is  a  verv  precise  apparatus 
which  is  used  in  large  laboratories.  Large  machines  for  testing 
eye-bars  have  been  built  by  bridge  companies,  and  numerous 
testing  laboratories  now  contain  apparatus  for  every  kind  of  work. 

The  capacity  of  a  testing  machine  is  the  tension  or  pressure 
which  it  can  exert.  A  small  machine  for  testing  cement  by  ten- 
sion has  usually  a  capacity  of  i  ooo  or  2  ooo  pounds.  Machines 
of  less  than  50  ooo  pounds  capacity  are  usually  operated  by  hand 
and  are  especially  useful  for  the  instruction  of  students;  while 
machines  of  higher  capacity  are  operated  by  power.  The  largest 
testing  machine  in  the  world  is  one  of  10  ooo  ooo  pounds  capacity 
at  Pittsburgh,  Pa.,  which  can  be  used  for  compression  only. 
A  list  of  the  large  testing  machines  in  the  United  States  is  given 
in  the  American  Civil  Engineers'  Pocket  Book. 

Fig.  184  shows  an  Olsen  testing  machine  of  40  ooo  pounds 
capacity.  The  power  is  applied  by  hand  by  means  of  the  crank 
on  the  left,  and  this  causes  the  four  vertical  screws  to  have  a  slow 
upward  or  downward  motion.  The  upper  ends  of  the  screws  are 
fastened  to  a  table  A,  which  hence  partakes  of  the  vertical  motion. 
When  a  tensile  test  is  made,  one  end  of  the  specimen  is  gripped 
by  jaws  in  the  movable  table  A  and  the  other  end  in  the  fixed 
table  B;  in  the  figure  a  tensile  specimen  is  seen  in  this  position. 
The  crank  is  then  turned  so  as  to  cause  the  movable  table  to 
descend  and  thus  a  tensile  load  is  brought  upon  the  specimen. 
This  load  is  weighed  on  the  lever  scale  at  the  right  by  moving 
the  weight  D  so  that  the  scale  arm  will  balance.  For  a  compressive 
test  the  specimen  is  placed  between  the  lower  fixed  table  C  and 


472 


TESTING  OF  MATERIALS 


CHAP.  XX 


the  movable  table  A,  the  latter  being  caused  to  descend  by  turning 
the  crank,  and  thus  a  compressive  load  is  brought  upon  the 
specimen. 

Tensile  tests  are  the  most  common,  and  some  commercial 
machines  are  arranged  with  an  autographic  recording  apparatus 
whereby  a  curve  is  drawn  which  shows  the  relation  between  the 
load  and  the  elongation  throughout  the  test.  There  are  also  a 


Fig.  184 

number  of  autographic  recording  devices  in  the  market,  which 
may  be  attached  to  any  machine.  When  such  a  graphic  record 
is  taken,  the  yield  point,  ultimate  strength,  and  ultimate  elonga- 
tion may  be  read  from  it.  Nearly  all  tensile  machines  may  be 
also  used  for  compressive  tests,  and  also  for  flexural  tests  on  short 
beams.  The  smaller  machines  are  operated  by  hand,  while 
power  is  required  to  run  the  larger  ones.  Screw  machines  in 
which  the  load  is  brought  upon  the  specimen  by  the  help  of  large 
screws  are  generally  preferred  in  the  United  States,  while  hydrau- 
lic machines  in  which  a  hydraulic  press  is  used  to  transmit  the 
load  to  the  specimen  are  preferred  in  Europe. 


ART.  184  TESTING  MACHINES  473 

Commercial  tests  of  materials  are  rarely  made  under  shearing 
and  torsional  stresses.  Thurston  in  1870  devised  a  torsion 
machine  for  small  specimens,  and  the  torsion  machines  of 
Olsen  and  of  Riehle,  which  are  found  in  the  laboratories  of  most 
engineering  colleges,  prove  very  serviceable  for  illustrating  the 
phenomena  of  twisting.  Impact  machines  have  been  built  for 
special  investigations,  but  the  only  one  on  the  market  is  that  of 
Keep,  which  is  designed  for  tests  of  bars  of  cast  iron.  Fatigue  or 
endurance  tests,  which  subject  the  specimens  to  alternating 
stresses  for  long  periods  of  time,  are  made  on  special  machines. 

The  testing  of  materials  has  assumed  such  great  importance 
since  1890  that  all  engineering  colleges  have  provided  laboratories 
for  the  purposes  of  instruction  and  research.  The  work  done 
by  some  of  these  has  proved  of  much  value  to  the  engineering 
profession;  the  work  of  Hatt  on  impact  tests  of  metals  and  that 
of  Talbot  on  reinforced-concrete  beams  may  be  cited  as  examples. 
Four  engineering  colleges  in  the  United  States  have  testing 
machines  of  600000  pounds  capacity  for  tension,  compression, 
and  flexure,  and  one  has  such  a  machine  of  800  ooo  pounds 
capacity,  while  one  has  a  machine  of  i  200  ooo  pounds  capacity 
for  compression  only. 

Large  testing  laboratories  have  also  been  established  by  the 
United  States  Government  in  its  arsenals,  and  since  1910  in  its 
Bureau  of  Standards.  The  valuable  work  done  by  Howard  at 
the  Watertown  Arsenal  and  for  the  Bureau  of  Standards  deserves 
special  mention  in  this  connection.  The  largest  testing  labor- 
atories, however,  are  found  in  Europe,  that  at  Berlin,  under  the 
directorship  of  Martens,  standing  at  the  head;  this  has  a  floor 
area  of  10  360  square  meters,  or  about  2j  acres. 

Prob.  184,  In  1912  a  brick  pier  4X4  feet  in  section  was  broken 
at  Pittsburgh  by  the  large  testing  machine  of  the  Bureau  of  Standards. 
Estimate  the  load  which  caused  failure.  Then  consult  the  engin- 
eering journals  of  September  or  October  of  1912  and  ascertain  the 
actual  load 


474  TESTING  OF  MATERIALS  CHAP.  XX 

ART.  185.    TEST  SPECIMENS 

Flat  specimens  are  used  to  test  the  tensile  strength  of  plates, 
and  Fig.  185<z  shows  the  standard  form  adopted  by  the  American 
Society  for  Testing  Materials.  The  thickness  of  the  specimen  is 
the  same  as  that  of  the  plate  from  which  it  is  cut;  its  total  length 
ranges  from  15  to  18  inches,  while  the  length  of  the  central  portion 
is  from  9  to  12  inches.  On  this  central  portion  prick  marks 
are  made  at  distances  i  inch  apart.  Then  after  rupture  the 


fAbout  3Z\ 

i  i 


—     1 

ylK 

&. 

-15  to  18- >{ 

Fig.  18oa 

distances  may  be  measured  again  in  order  to  ascertain  the  per- 
centage of  elongation. 

Round  specimens  are  used  for  the  tensile  tests  of  most  mate- 
rials other  than  plates.  Wires  and  rods  are  usually  tested  just 
as  they  come  from  the  mill,  the  length  between  the  jaws  of  the 
machine  being  more  than  ten  times  the  diameter.  Standard 
round  specimens  are  cut  from  axles,  beams,  shafts,  and  other 
manufactured  products,  these  being  turned  in  a  lathe  and  screws 
cut  on  the  enlarged  ends.  Prior  to  1895  the  standard  diameter 
of  the  round  specimen  was  i  inch,  and  its  standard  length  was 
8  inches,  this  being  the  distance  between  two  marks  which  are 
placed  on  the  central  portion  for  the  purpose  of  measuring  the 
elongations.  After  1900,  however,  there  came  into  use  a  smaller 
size  which  is  now  widely  used,  although  the  8-inch  specimen  is 
still  required  for  some  classes  of  work.  This  smaller  size  is  called 
the  2 -inch  specimen,  because  the  central  part  is  a  little  more  than 
2  inches  long  and  the  marks  are  placed  upon  it  2  inches  apart; 
the  diameter  of  this  specimen  is  0.5  or  sometimes  0.505  inches. 


ART.  185 


TEST  SPECIMENS 


475 


In  Fig.  1855  the  upper  specimen  has  not  been  tested,  while  the 
two  lower  ones  have  been  broken;  see  Art.  12  for  an  account 
of  these  tests. 


Fig.  185c 

Specimens  for  compressive  tests  of  metals  are  usually  cylin- 
ders, while  cubes  are  mostly  used  for  cement,  concrete,  and  stone, 
and  rectangular  prisms  for  timber.  The  length  of  a  cylinder 
or  prism  should  be  between  2-J-  and  4  times  its  diameter  or  side, 


£76  TESTING  OF  MATERIALS  CHAP.  XX 

in  order  that  there  may  be  opportunity  for  the  oblique  shearing 
to  properly  occur.  Fig.  lS5c  shows  a  cube  of  neat  cement  2  inches 
square  which  has  failed  under  compression,  and  also  a  timber 
prism  2X2  inches  in  section  area,  where  the  oblique  shearing 
occurs  far  more  satisfactorily  than  in  the  cube.  This  method 
of  failure  is  characteristic  of  brittle  materials  (Art.  18).  Com- 
pressive  tests  are  rarely  used  for  metals  on  account  of  the  expense 
of  making  the  specimens  with  ends  truly  parallel. 

Specimens  for  flexural  tests  of  cast  iron  are  usually  i  inch 
square  and  14  inches  long,  which  are  placed  on  supports  12  inches 
apart  and  broken  by  a  load  at  the  middle  of  the  span.  Specimens 
ij  inches  in  diameter  and  15  inches  in  span  are  also  used.  Foi 
other  materials  no  standard  sizes  have  yet  been  adopted. 

Prob.  185.  Wherfthe  standard  square  bar  is  used  for  a  flexural  test 
of  cast  iron,  show  that  the  modulus  of  rupture  can  be  found  by 
multiplying  the  breaking  load  in  pounds  by  18. 

ART.  186.    TENSILE  TESTS 

Flat  specimens  are  usually  gripped  by  the  jaws  of  the  machine; 
while  the  standard  round  specimens  screw  into  nuts  to  whicl' 
the  tension  is  applied.  The  load  transmitted  through  the  speci- 
men is  weighed  by  a  scale  at  the  end  of  a  compound  lever.  In 
commercial  tests  the  ultimate  elongation  is  alone  measured; 
this  is  done  by  two  marks  on  the  specimen  and  measuring  the 
distance  between  them  before  and  after  rupture.  In  scientific 
tests  an  extensimeter  is  attached  to  the  specimen,  so  that  the 
elongation  can  be  read  at  each  increment  of  weight.  The  elonga- 
tion is  usually  expressed  as  a  percentage  of  the  original  length 
between  the  two  marks  on  the  specimen,  and  it  is  always 
desirable  that  the  original  length  and  the  diameter  of  specimen 
should  be  mentioned  in  the  report,  since  it  will  not  be  the  same 
for  the  2-inch  as  for  the  8-inch  specimen.  For  ductile  materials, 
like  wrought  iron  and  mild  steel,  it  is  customary  to  slowly  reduce 
the  load  after  the  highest  value  is  reached;  the  material  is  then 
flowing  rapidly,  so  that  the  elongation  continues  to  increase, 
and  hence  a  greater  percentage  of  elongation  is  obtained. 


ART.  186  TENSILE  TESTS  477 

When  it  is  specified  that  certain  methods  of  testing  or  certain 
test  specimens  are  to  be  used,  these  should  be  followed  most 
carefully.  Full  specifications  are,  however,  rarely  made  by  a 
purchaser,  so  that  considerable  latitude  is  allowed,  but  in  all 
cases  it  should  be  the  aim  of  the  engineer  in  charge  of  the  tests 
to  so  conduct  the  work  that  it  shall  be  well  and  truly  done. 
Certain  general  rules  regarding  testing  machines  and  their  use 
will  hence  be  given  here,  and  the  observance  of  these  as  far  as 
practicable  will  conduce  to  uniformity  of  methods  and  to  reliability 
of  the  results . 

1.  A  machine  should  be  so  constructed  that  the  load 
borne  by  the  specimen  alone  is  registered  upon  the  weighing 
scale,    so    that    its    readings    may    not    include    any  force 
expended  in  friction  on  the  pivots  or  moving  parts  of  the 
machine. 

2.  A  machine  should  be  from  time  to  time  rated  or  cali- 
brated, to  ascertain  if  the  readings  of  the  weighing  scale 
are  correct,  or  that  the  errors  of  its  readings  may  be  known. 

3.  The  construction  and  operation  of  the  machine  should 
be  such  that  the  specimens  may  not  be  subject  to  shock. 

4.  The  holders  or  jaws  which  gripe  the  ends  of  a  tensile 
specimen  should    be  so  arranged   that   the   resultant   load 
coincides  with  the  axis  of  the  specimen,  in  order  that  the 
stress  may  be  uniformly  distributed  over  the  section  area. 

5.  The  use  of  serrated  wedges  for  holding  the  ends  of 
specimens  is  not  advisable  unless  those  ends  are  larger  in 
section  area  than  the  main  part  of  the  specimen. 

6.  The  load  upon  specimens  of  ductile  materials  should 
be  applied  at  a  slower  rate  within  the  elastic  limit  than  after 
that  limit  is  passed. 

7.  When  the  elastic  limit  is  specified,  this  is  not  to  be 
determined  by  the  drop  of  the  scale  beam,  but  by  measuring 
the  increments  of  elongation  corresponding  to  increments 
of  the  load;  the  drop  of  the  beam  indicates  the  yield  point. 

8.  Percentages  of  elongation  should  be  accompanied  by 
a  statement  of  the  size  of  the  test  pieces  from  which  they 
were  determined. 


478  TESTING  OF  MATERIALS  CHAP.  XX 

The  above  rules  are  quite  general,  but  they  seem  to  be 
essential  ones  for  the  proper  conduct  of  tensile  tests,  not  only 
for  metals,  but  also  for  timber  and  cement.  For  detailed  rules 
regarding  metals,  see  the  l  Specifications  for  Standard  Methods 
of  Testing '  adopted  by  the  American  Society  for  Testing  Mate- 
rials which  are  published  in  its  annual  Year  Book.  These  speci- 
fications give  also  methods  of  calibrating  testing  machines. 

Prob.  186.  Consult  the  Year  Book  above  mentioned  and  ascer- 
tain how  the  calibration  of  a  testing  machine  is  made. 

ART.  187.     COMPRESSIVE  TESTS 

Rules  i,  2,  3  of  the  preceding  article  apply  also  here.  The 
following  additional  rules  for  compressive  tests  of  metals  are  taken 
from  the  standard  specification  above  cited: 

4-  The  test  specimen  shall  be  a  cylinder  having  plane 
ends  truly  normal  to  its  axis.  The  diameter  of  the  specimen 
shall  not  be  less  than  i  inch  nor  greater  than  1.13  inches, 
the  former  being  preferred.  Its  length  shall  be  between  2.5 
and  4  diameters.  No  bedding  should  be  used  for  the  ends 
of  the  specimen. 

5.  The  bearing  blocks  which  transmit  the  pressure  from 
the  testing  machine  should  be  truly  normal  to  the  plane 
ends  of  the  specimen.     To  secure  this,  one  of  the  blocks 
should   be  provided   with  a  hemispherical   bearing  which 
can  turn  freely. 

6.  The    speed   of   compression    should    be    slow,    not 
exceeding    one-tenth   of  an  inch  per  minute.      Near  the 
elastic  limit  and  yield  point  the  load  should  be  increased 
very  slowly. 

7.  For    determining    modulus  of  elasticity,    the  linear 
compression  of  the  specimen  should  be  observed  by  a  pre- 
cise compressometer  which  is  attached  to  the  specimen  and 
does  not  touch  the  bearing  blocks  of  the  machine.     Read- 
ings of  the  compressometer  should  be  taken  for  three  loads, 
the  first  at  about  one-fourth,  the  second  at  about  one-half, 
and  the  third  at  about  three-fourths  of  the  elastic  limit. 


ART.  188  MISCELLANEOUS   TESTS  479 

8.  To  determine  the  elastic  limit,   several  readings  of 
the   compressometer  should  be  taken,  as  that  limit  is  ap- 
proached, for  load  increments  of  i  ooo  pounds  per  square 
inch. 

9.  The  yield  point  is  to  be  noted  as  corresponding  to 
that   load    for  which  the   compressometer  shows  a  linear 
compression  without  an  increase  in  load.     In  the  absence 
of  a  compressometer  this  point  may  be  noted,  for  ductile 
materials,  by  the  drop  of  the  scale  beam. 

10.  Measurements  for  the  modulus  of  elasticity,  elastic 
limit,  and  yield  point  may  be  made,  if  desired,  on  a  specimen 
ranging  in   length   from    10  to   15  diameters,  since  it  may 
often  be  difficult  to  apply  a  compressometer  in  a   length 
shorter  than  4  inches. 

1 1.  The  record  of  the  test  should  mention  any  phenomena 
observed  near  the  elastic  limit  and  yield  point.    The  manner 
of  final  failure  should  also  be  noted  when  the  test  is  carried 
to  this  limit. 

For  stone,  concrete,  and  timber  no  standard  methods  for  com- 
pressive  tests  have  yet  been  adopted.  The  general  principles 
given  in  the  above  rules  are  applicable,  however,  and  the  general 
practice  regarding  test  specimens  is  stated  in  Art.  185. 

Prob.  187.  Leather,  paper,  or  plaster,  has  often  been  used  between 
the  ends  of  the  specimen  and  the  bearing  blocks.  Why  should  this 
not  be  done? 

ART.  188.    MISCELLANEOUS  TESTS 

Flexural  tests  are  not  made  in  common  commercial  work, 
except  for  cast  iron,  but  tests  are  frequently  made  for  determining 
constants  to  be  used  in  designing.  In  such  cases  it  is  best  that 
the  specimen  should  be  a  model  of  the  actual  piece  which  is  to 
be  designed.  Thus,  if  it  is  desired  to  know  the  modulus  of 
rupture  for  a  timber  beam  2X3X24  feet,  a  breaking  test  on  a 
specimen  2X3X24  inches  will  give  a  reliable  result.  The  load 
is  usually  applied  at  the  middle  of  a  beam  supported  at  its  ends, 
or  at  two  points  which  divide  the  span  into  three  equal  parts. 
When  the  deflection  is  observed  the  apparatus  should  not  be 


480 


TESTING  OF  MATERIALS 


CHAP.  XX 


Fig.  188 


ART.  188  MISCELLANEOUS  TESTS  481 

attached  to  the  supports,  for  their  compression  might  affect  the 
observed  values.  When  the  elastic  limit  is  not  exceeded  the 
modulus  of  elasticity  of  the  material  may  be  computed  from  the 
observed  deflection. 

Torsion  tests  are  not  used  for  commercial  purposes,  except 
for  wire,  but  in  laboratory  work  much  has  been  learned  by  the 
use  of  torsion  machines.  Fig.  188  shows  three  specimens  which 
have  been  subjected  to  torsion,  the  first  being  a  square  steel  bar 
which  broke  at  the  upper  end,  the  second  a  round  bar  of  cast 
iron,  and  the  third  a  rectangular  steel  bar  ruled  with  white  lines 
in  order  that  the  distortions  might  be  studied  *  discussions  of  these 
experiments  are  given  in  Arts.  94  and  99. 

Impact  tests  are  regarded  as  of  much  value  in  judging  of  the 
quality  of  ductile  metals.  The  cold-bend  test,  briefly  described 
in  Art.  24,  is  one  that  has  long  been  used  in  all  mills  where  wrought 
iron  or  steel  is  produced.  The  bending  of  the  specimen  is 
generally  done  by  blows  of  a  hammer,  although  steady  pressure 
is  sometimes  employed.  Notwithstanding  that  no  numerical 
results,  except  the  final  angle  of  bending,  are  obtained  from  the 
cold-bend  test,  the  general  information  that  it  gives  is  of  the 
highest  importance,  so  that  is  had  been  said  that,  if  all  tests  of 
metals  except  one  were  to  be  abandoned,  the  cold-bend  test 
should  be  the  one  to  be  retained.  In  the  rolling-mill  it  is  used 
to  judge  of  the  purity  and  quality  of  the  muck  bar;  in  the  steel 
mill  it  serves  to  classify  and  grade  the  material  almost  as  well  as 
chemical  analysis  can  do,  and  in  the  purchase  of  shape  iron  it 
affords  a  quick  and  satisfactory  method  of  estimating  toughness, 
ductility,  strength,  and  capacity  to  resist  external  work. 

Hardness  tests  of  metals  may  be  made  by  forcing  a  hardened 
steel  ball  into  the  specimen  under  static  pressure  and  measuring 
the  depth  of  the  indentation.  The  total  load  divided  by  the 
area  of  the  indented  spherical  cup  is  called  the  hardness  number, 
the  greater  the  number  the  higher  the  hardness  of  the  material. 
Another  method  is  by  dropping  a  diamond-pointed  plunger 
through  a  glass  tube  onto  the  specimen,  the  height  of  rebound  of 
the  specimen  indicating  the  hardness  of  the  specimen. 


482  TESTING  OP  MATERIALS  CHAP.  XX 

For  fuller  information  regarding  testing  machines  and  methods 
of  testing  see  American  Civil  Engineers'  Pocket  Book,  and  the 
Year  Books  of  the  American  Society  for  Testing  Materials. 

Prob.  188.  Consult  American  Civil  Engineers'  Pocket  Book  and 
describe  (a)  a  shearing  test  for  metals,  (b)  the  micrometer  extenso- 
meter,  (c)  the  tensile  test  for  cement. 

ART.  189.     SPECIFICATIONS  FOR  STRUCTURAL  STEEL 

When  materials  are  to  be  purchased,  a  set  of  rules  is  usually 
prepared  giving  requirements  regarding  quality  and  tests,  and 
these  rules  are  called  '  specifications.'  There  is  much  variation 
in  such  specifications  owing  to  the  different  opinions  of  buyers 
and  the  use  which  is  to  be  made  of  the  material.  The  following 
specifications  for  structural  steel  will  give  the  student  an  idea 
of  the  extent  and  scope  of  the  requirements  which  are  generally 
demanded  for  steel  to  be  used  in  buildings,  bridges,  and  ships. 
The  specifications  which  have  been  adopted  by  the  American 
Society  for  Testing  Materials  and  the  American  Railway  Engin- 
eering and  Maintenance  of  Way  Association  differ  from  these 
only  in  a  few  particulars:' 

1.  Structural  steel  shall  be  made  by  the  open-hearth  process. 

2.  Each  of  the  three  classes  of  structural  steel  shall  conform  to 
the  following  limits  in  chemical  composition:  Sulphur  shall  not  ex- 
ceed 0.06  percent;  Phosphorus  shall  not  exceed  0.07  percent  when 
the  steel  is  made  by  the  acid  process,  and  not  exceed  0.04  percent 
when  it  is  made  by  the  basic  process. 

3.  There  shall  be  three  classes  of  structural  steel  for  bridges  and 
ships,  namely,  rivet  steel,  soft  steel,  and  medium  steel,  which  shall 
conform  to  the  following  physical   requirements:    Rivet   steel   shall 
range  in  tensile  strength  from  50000  to  55  ooo  pounds  per  square 
inch,  have  a  yield  point  not  less  than  the  33  ooo  pounds  per  square 
inch,  and  the  elongation  in  8  inches  shall  not  be  less  than  26  per- 
cent.   Soft  steel  shall  range  in  tensile  strength  from  55  ooo  to  60000 
pounds  per  square  inch,    have  a  yield   point  not   less   than  35  ooo 
pounds  per  square  inch,  and  the  elongation  in  8  inches  shall  not  be 
less  than  25  percent.      Medium  steel  shall  range  in  tensile  strength 


ART.  189         SPECIFICATIONS   FOR  STRUCTURAL  STEEL  483 

from  60000  to  65000  pounds  per  square  inch,  have  a  yield  point 
not  less  than  37  ooo  pounds  per  square  inch,  and  the  elongation  in 
8  inches  shall  not  be  less  than  23  percent. 

4.  For  each  increase  of  J  inch  in  a  flat  specimen  above  a  thick- 
ness of  |  inches,    a  deduction  of  i  shall  be  made  from  the  above 
specified  elongation.      For  each  decrease  of  ^  inch  below  a  thick- 
ness of  •£$  inches,  a  deduction  of  2\  shall  be   made  from  the  above 
specified  elongation.     For  bridge  pins  the  required  elongation  shall 
be  5  less  than  that  above  specified,  as  determined  on  a  test  specimen 
the  center  of  which  shall  be  one  inch  from  the  surface  of  the  pin. 

5.  Eye-bars  shall  be  of  medium  steel.      Full-sized  tests   shall 
show  12 \  percent  elongation  in  15  feet  of  the  body  of  the  eye-bar, 
and  the  tensile  strength  shall   not  be  less  than  55  ooo  pounds  per 
square  inch.     Eye-bars  shall  be  required  to  break  in  the  body,  but 
should  an  eye-bar  break  in  the  head,  and  show  i2|  percent  elonga- 
tion in  15  feet  and  the  tensile  strength  specified,  it  shall  not  be  cause 
for  rejection,    provided  that  not  more  than  one-third  of  the  total 
number  of  eye-bars  tested  break  in  the  head. 

6.  The  three  classes  of  structural  steel  shall  conform  to  the  lul- 
lowing  bending  tests;  and  for  this  purpose  the  test  specimen  shall  be 
ij  inches  wide,  if  possible,  and  for  all  material  f  inches  or  less  in 
thickness  the  test  specimen  shall  be  of  the  same  thickness  as  that  of 
the  finished  material  from  which  it  is  cut,  but  for  material  more  than 
\  inches  thick  the  bending-test  specimen  may  be  \  inch  thick: 

Rivet  steel  shall  bend  cold  180  degrees  flat  on  itself 
without  fracture  on  the  outside  of  the  bent  portion.  Rivet 
rounds  shall  be  tested  of  full  size  as  rolled. 

Soft  steel  shall  bend  cold  180  degrees  flat  on  itself  with- 
out fracture  on  the  outside  of  the  bent  portion. 

Medium  steel  shall  bend  cold  180  degrees  around  a 
diameter  equal  to  the  thickness  of  the  specimen  tested,  with- 
out fracture  on  the  outside  of  the  bent  portion. 

7.  The  standard  test  specimen  of  8  inches  gauged  length  shall  be 
used  to  determine  the  physical  properties   specified  in  paragraphs 
Nos.   3  and  4.     The  standard  size  of  the  test  specimen  for  sheared 
plates  shall  be  i  J  inches  in  width  for  a  length  not  less  than  9  inches, 
and  of  a  thickness   equal  to  that  of  the  plate.     For  other  material 


484  TESTING  OF  MATERIALS  CHAP.  XX 

the  test  specimen  may  be  the  same  as  for  sheared  plates  or  it  may  be 
planed  or  turned  parallel  throughout  its  entire  length,  and  in  all 
cases  where  possible  two  opposite  sides  of  the  test  specimen  shall 
be  the  rolled  surfaces.  Rivet  rounds  and  small  rolled  bars  shall  be 
tested  of  full  size  as  rolled. 

8.  One  tensile  test   specimen   shall   be  taken   from  the  finished 
material   of  each   melt   or  blow,  but   in  case  this  develops  flaws,  or 
breaks  outside   of  the  middle   third  of  its  gauged  length,  it  may  be 
discarded  and  another  test  specimen  substituted  therefor. 

9.  One   test    specimen    for    bending    shall  be    taken    from    the 
finished  material  of  each   melt   as  it  comes  from  the  rolls,  and  for 
material   J  inches   and    less  in  thickness,    this   specimen  shall  have 
the  natural  rolled  surface  on  two  opposite  sides.     The  bending-test 
specimen   shall    be    ij   inches   wide,    if  possible,    and   for  material 
more  than  J  inches  thick,  the  bending-test  specimen  may  be  \  inch 
thick.     The  sheared  edges  of  bending-test  specimens  may  be  milled 
or  planed. 

10.  Material  which   is  to  be   used   without  annealing  or  further 
treatment   shall   be  tested  for  tensile   strength  in  the  condition  in 
which    it    comes    from    the    rolls.      Where    it    is    impracticable    to 
secure    a    test    specimen    from   material   which   has    been   annealed 
or   otherwise    treated,    a  full-sized  section  of  tensile-test  specimen 
length    shall   be    similarly    treated    before   cutting    the    tensile-test 
specimen  therefrom. 

11.  For  the  purpose  of  this  specification,  the  yield  point  shall  be 
determined  by  careful  observation  of  the  drop  of  the  beam  or  halt 
in  the  gauge  of  the  testing  machine. 

12.  In   order   to    determine    if  the    material    conforms    to   the 
chemical  limitations  prescribed  in  paragraph  No.  2  herein,  analysis 
shall  be  made  of  drillings  taken  from  a  small  test  ingot. 

13.  The   variation   in  cross-section   or   weight   of  more  than  2\ 
percent   from   that   specified   will   be   sufficient  cause   for  rejection, 
except  in  the  case   of  sheared   plates,  which  will  be  covered  by  the 
following  permissible  variations: 

Plates  i2j  pounds  per  square  foot,  or  heavier,  up  to  100  inches 
wide,  when  ordered  to  weight,  shall  not  average  more  than  2j  per- 


ART.  189  SPECIFICATIONS  FOR  STRUCTURAL  STEEL  485 

cent  variation  above  or  below  the  theoretical  weight.  When  100 
inches  wide  and  over,  5  percent  above  or  5  percent  below  the 
theoretical  weight. 

Plates  under  i2|  pounds  per  square  foot,  when  ordered  to 
weight,  shall  not  average  a  greater  variation  than  the  following: 
When  less  than  75  inches  wide,  2j  percent  above  or  below  the 
theoretical  weight.  When  75  inches  wide  up  to  zoo  inches  wide, 
5  percent  above  or  3  percent  below  the  theoretical  weight.  When 
100  inches  wide  and  over,  10  per  cent  above  or  3  per  cent  below  the 
theoretical  weight. 

When  plates  are  ordered  to  gauge,  a  variation  of  more  than 
r^j  inch  below  that  specified  for  any  dimension  will  be  sufficient 
cause  for  rejection.  An  excess  in  weight  above  the  nominal  weight 
may,  however,  be  allowed,  as  agreed  upon  between  the  inspector 
and  the  manufacturer, 

14.  Finished  material  must  be  free  from  injurious  seams,  flaws, 
defective  edges,  or  cracks,  and  have  a  workmanlike  finish. 

15.  Every  finished  piece  of  steel  shall  be  stamped  with  the  melt 
number,  and  steel  for  pins  shall  have  a  melt  number  stamped  on  the 
ends.     Rivets  and  lacing  steel,  and  small  pieces  for  pin  plates  and 
stiffeners,  may  be  shipped  in  bundles,  securely  wired  together,  with 
the  melt  number  on  a  metal  tag  attached. 

1 6.  The    inspector   representing   the    purchaser,    shall    have    all 
reasonable  facilities  afforded  to  him  by  the  manufacturer  to  satisfy 
him  that  the  finished  material  is  furnished  in  accordance  with  these 
specifications.     All  tests  and  inspections  shall  be  made  at  the  place 
of  manufacture,  prior  to  shipment. 

Prob.  189#.  Consult  Proceedings  of  the  American  Society  for 
Testing  Materials,  Vol.  IV,  1904,  and  ascertain  the  tests  recom- 
mended by  Webster  for  detecting  brittle  steel. 

Prob.  189$.  Consult  the  Year  Book  of  the  American  Society  for 
Testing  Materials  and  describe  the  toughness  test  for  rock  used  in 
making  macadam  roads. 

Prob.  189r.  Consult  a  paper  by  Goss  in  Vol.  Ill  of  Proceedings 
of  American  Society  for  Testing  Materials,  and  describe  the  drop 
testing  machine  of  the  Master  Car  Builders'  Association. 


486  TESTING  OF  MATERIALS  CHAP.  XX 


ART.  190.    UNIFORMITY  IN  TESTING 

It  is  very  important  in  order  that  the  results  of  tests  made 
in  different  laboratories  may  be  compared  and  correct  conclu- 
sions be  drawn  therefrom,  that  the  methods  of  testing  should  be 
uniform.  In  1882  a  number  of  German  professors  met  at  Munich 
to  discuss  this  question,  and  other  conferences  were  held  in  later 
years  at  which  engineers  from  other  European  countries  were 
present.  In  1895  tne  International  Association  for  Testing  Mate- 
rials was  formally  organized  at  Zurich,  its  object  being  '  The  devel- 
opment and  unification  of  standard  methods  of  testing  for  the 
determination  of  the  properties  of  materials,  and  also  the  per- 
fection of  apparatus  for  that  purpose.'  International  congresses 
have  been  held  since  at  intervals  of  two  or  three  years,  the  sixth 
congress  being  at  New  York  in  September,  1912,  which  was 
attended  by  180  official  delegates  representing  twenty  different 
countries,  and  by  about  600  other  members  of  the  International 
Association.  The  total  number  of  members  of  this  Association 
in  Jan.,  1913,  was  2  964,  of  whom  622  were  in  the  United  States. 

The  American  Section  of  the  International  Association, 
organized  in  1898,  became  in  1901  the  American  Society  for  Testing 
Materials.  In  1913  there  were  over  i  500  members  of  this  Society, 
of  whom  about  600  were  also  members  of  the  Association.  The 
Society  publishes  an  annual  volume  of  Proceedings  and  also  a 
Year  Book  containing  specifications.  Its  sixteenth  annual  meet- 
ing was  held  in  June,  1913.  The  specifications  issued  by  the 
American  Society  for  Testing  Materials  have  been  prepared 
by  committees  on  which  both  producers  and  consumers  were 
represented,  and  they  are  revised  from  time  to  time  as  the  science 
and  art  of  the  subject  of  materials  advances. 

In  conclusion  it  may  be  noted  that  all  the  theoretic  discussions 
of  this  volume  are  of  value  in  writing  specifications  for  materials, 
as  well  as  in  conducting  tests  in  the  laboratory.  When  tensile 
or  compressive  tests  are  made  it  must  not  be  forgotten  that  reliable 
results  cannot  be  obtained  unless  the  spgcimen  is  placed  in  the 


ART.  190  UNIFORMITY  IN  TESTING  ?     487 

machine  in  such  a  manner  that  the  load  is  truly  centered;  when 
a  beam  is  to  be  tested  the  theory  of  flexure  must  be  understood. 
Clear  and  definite  ideas  regarding  elastic  limit  and  yield  point 
will  avoid  misunderstandings.  The  development  of  shearing 
stresses  in  a  compressive  test  should  not  be  overlooked.  The 
true  stresses  as  distinguished  from  the  apparent  ones  must  be 
kept  in  mind.  Theory  and  practice  should  always  go  hand  in 
hand,  each  aiding  and  supplementing  the  other. 

Prob.  190#.  Read  the  address  of  President  Charles  B.  Dudley,  on 
che  duties  and  responsibilities  of  the  inspecting  engineer,  given  in 
1905  at  the  annual  meeting  of  the  American  Society  for  Testing 
Materials. 

Prob.  190&  Refer  to  Hatt  and  Marburg's  Bibliography  of  Impact 
and  Testing  Machines  in  Vol.  II  of  Proceedings  of  American  Society 
of  Testing  Materials,  and  describe  one  or  more  machines  for  making 
impact  tests. 


488  APPENDIX  AND  TABLES 


APPENDIX    AND   TABLES 

ART.  191.     VELOCITY  OF  STRESS 

WHEN  an  external  force  is  suddenly  applied  to  a  body,  the 
stresses  produced  are  not  instantaneously  generated,  but  are 
propagated  by  a  wave-like  motion  through  the  mass.  Hence 
there  is  a  velocity  of  transmission  of  stress  which  will  be  shown 
to  depend  upon  the  stiffness  and  density  of  the  material.  In 
fact,  a  sudden  stress  is  propagated  through  a  body  in  the  same 
manner  as  sound  is  propagated  through  air  or  water.  Let  v 
be  this  velocity,  w  the  weight  of  the  material  per  cubic  unit  at 
a  place  where  the  acceleration  of  gravity  is  g,  and  E  the  modulus 
of  elasticity  of  the  material.  It  is  required  to  find  v  in  terms 
of  w,  g,  and  E. 

If  F  is  a  force  which  acting  continuously  for  one  second  upon 
a  body  of  the  weight  W  produces  the  velocity  u,  and  if  the  same 
body  when  falling  vertically  acquires  under  the  action  of  gravity 
the  velocity  g  in  one  second,  then  the  constant  forces  are  pro- 
portional to  the  accelerations  that  they  produce;  hence, 

F/W=u/g  or  Fg=Wu 

which  is  one  of  the  well-known  laws  of  mechanics. 

Now  let  a  unit-stress  5  be  applied  to  the  end  of  a  bar  of  section- 
area  unity,  producing  the  unit- elongation  s  upon  the  first  element 
of  its  length.  The  elongation  of  the  first  element  transmits  the 
stress  to  the  second  element,  and  this  in  turn  produces  an  elonga- 
tion of  the  second  element,  and  so  on.  At  the  end  of  one  second 
of  time  the  length  v  is  stressed,  and  the  total  elongation  in  that 
length  will  be  EV.  Thus  in  one  second  the  center 'of  gravity 
of  the  bar  is  moved  the  distance  %ev,  and  its  velocity  u  at  the 
end  of  the  second  is  ev.  Now  referring  to  the  formula  Fg  =  Wu, 
the  value  of  F  is  S  which  is  equal  to  eE,  the  value  of  W  is  wv, 
and  hence, 

eE.g  =  wv.ev  or  v 


ART.  191  VELOCITY  OF  STRESS  489 

which  is  the  formula  for  the  velocity  of  wave  propagation  in 
elastic  materials  first  deduced  by  Newton. 

Taking  for  g  the  mean  value  32.16  feet  per  second  per  second, 
for  w  the  values  in  Table  1,  for  E  the  values  in  Table  2,  and 
reducing  E  to  pounds  per  square  foot,  since  g  and  w  are  in  terms 
of  feet,  the  mean  values  of  the  velocity  of  transmission  of  stress 
in  different  materials  are  found  to  be, 

for  Timber,  v=  13  200  feet  per  second 

for  Stone,  v=  13  200  feet  per  second 

for  Cast  Iron,  v=i2  400  feet  per  second 

for  Wrought  Iron,  v=  15  500  feet  per  second 

for  Steel,  v=  17  200  feet  per  second 

For  water  confined  in  a  pipe,  the  value  of  E  is  the  same  as  the 
volumetric  modulus  G  (Art.  182),  and  taking  w  as  62 \  pounds 
per  cubic  foot,  the  velocity  v  is  about  4  670  feet  per  second,  which 
agrees  well  with  experiments  on  the  velocity  of  sound  in  water. 

In  the  mathematical  theory  of  elasticity,  the  velocity  of  trans- 
mission of  stress  must  be  taken  into  account  in  order  to  obtain 
complete  solutions  of  the  problems  of  impact  and  suddenly 
applied  forces.  The  above  formula  also  gives  the  velocity  of 
sound,  light,  and  all  wave  propagations  in  elastic  media.  The 
ratio  w/g  is  a  constant  for  the  same  material  at  any  point 
in  space,  and  it  expresses  the  density,  while  E  is  an  index  of 
the  stiffness.  At  the  surface  of  the  earth  the  quantity  E/w 
for  steel  is  about  8  820  coo,  but  for  the  ether  it  must  be  about 
30  ico  coo  ooo  ooo  ooo  in  order  to  account  for  the  fact  that  the 
velocity  of  light  is  984  ooo  ooo  feet  per  second.  The  stiffness  of 
the  ether  is  hence  very  great  compared  to  its  density;  if  its  density 
be  one  one-thousandth  of  that  of  hydrogen,  its  stiffness  is  37 
times  as  great  as  that  of  steel.  The  opinion  has  long  prevailed 
that  the  force  of  gravitation  is  instantaneously  propagated  through 
the  ether,  but  the  indications  now  are  that  its  velocity  is  the 
same  as  that  of  light  and  electricity. 

Prob.  191.  Verify  the  statements  in  the  last  paragraph.  Consult 
Van  Nostrand's  Science  Series,  No.  85,  and  ascertain  the  values  deduced 
by  Wood  lor  the  density  and  stiffness  of  the  ether  of  space. 


490  APPENDIX  AND  TABLES 


ART.  192.     ELASTIC-ELECTRIC  ANALOGIES 

In  Art.  193  of  Treatise  on  Hydraulics  there  are  pointed  out 
some  of  the  analogies  between  hydraulic  and  electric  phenomena. 
The  theory  of  elasticity  furnishes  other  analogies  which  are 
interesting  and  one  of  these  is  perhaps  more  perfect,  from  a 
formal  point  of  view,  than  any  that  is  furnished  by  hydraulics. 
Let  a  bar  of  length  /  and  section  area  a  be  under  the  axial  tension 
P,  let  E  be  the  modulus  of  elasticity  of  the  material  and  e  the 
change  of  length  due  to  the  tension;  then  from  Art.  10,  this  elon- 
gation is  e  =  (l/aE)P.  Let  the  reciprocal  of  E  be  called  E',  then 
e  =  E'(l/a)P.  This  equation  is  the  same  as  that  given  by  Ohm's 
law  for  the  loss  in  voltage  when  a  current  flows  through  a  wire, 
if  e  represents  the  lost  voltage  and  P  the  current;  the  quantity 
E'(l/a)  is  called  electric  resistance  and  it  varies  directly  as  the 
length  of  the  wire,  inversely  as  its  section  area,  and  directly  as 
the  specific  electric  resistance  E' '.  The  formal  analogy  is  per- 
fect, but  the  fundamental  ideas  are  quite  different  in  the  two 
cases.  In  the  bar  e  is  the  loss  in  length  which  varies  directly 
as  /,  inversely  as  a,  and  directly  as  the  reciprocal  of  E,  but  the 
phenomenon  is  a  static  one  entirely,  for  no  energy  is  lost  or 
transmitted  through  the  bar. 

Referring  now  to  Fig.  1100,  let  /i,  I2j  h  be  the  lengths  of  the 
three  columns  under  the  compression  P,  and  a\,  a2,  a3  their 
section  areas,  while  'Ei,  E2>  £3  are  their  moduluses  of  elasticity. 
Let  the  quantities  /i/fliEi,  l2/a2E2,  /s/^s-Ea  be  called  resistances 
and  be  designated  by  the  letters  rb  r2,  r3.  Then  by  formula  (110) 
the  total  change  of  length  of  the  compound  column  is  given  by 
e=(r\  +  r2  +  r3)P.  This  may  be  called  an  arrangement  'in 
series,'  and  as  in  electric  flow,  the  total  resistance  is  the  sum 
of  the  separate  resistances. 

Fig.  1106  represents  a  column  which  may  be  said  to  be  an 
arrangement  '  in  parallel,'  and  here  the  three  lengths  are  equal 
to  /,  while  the  section  areas  are  #1,  a2,  a3.  Formula  (110)"  gives 
the  change  of  length  for  this  case  as  e  =  P/(i/ri  +  i/r2  +  i/r3), 
which  agrees  with  the  electric  law  governing  the  loss  of  voltage 


ART.  193.  MISCELLANEOUS    PROBLEMS  491 

in  a  branched  circuit,  the  total  resistance  being  the  reciprocal  of 
the  sum  of  the  reciprocals  of  the  separate  resistances.  The 
loads  PI,  P2,  P^  on  the  three  parts  of  the  column  correspond 
to  the  currents  in  the  three  parts  of  the  divided  wire,  while  the 
change  of  length  e  corresponds  to  the  drop  in  voltage  which  is 
the  same  for  each  of  the  three  parts.  The  load  P  divides  between 
the  three  parts  inversely  as  their  resistances  to  change  of  length, 
while  the  electric  current  divides  between  three  branches  in- 
versely as  their  electric  resistances. 

With  respect  to  work  the  analogy  is  less  perfect.  Resuming 
the  equation  e  =  (l/aE)P  for  the  change  in  length  of  a  bar  under 
axial  stress,  the  product  of  P  and  e  is  work,  while  this  product 
is  work  per  unit  of  time  if  e  represents  voltage  and  P  represents 
current.  This  is  because  P  represents  force  for  the  bar,  while 
for  the  electric  wire  it  represents  electric  charge  per  unit  of  time. 
For  the  bar  the  external  work  is  spent  in  storing  internal  energy 
in  the  bar;  for  the  wire  the  work  Pe  is  lost  in  heat.  The  above 
analogy  is  hence  merely  a  formal  one  and  the  fundamental  ideas 
are  quite  different  in  the  two  cases. 

The  phenomena  of  torsion  afford  another  analogy  and  here 
energy  may  be  transmitted  through  a  rotating  shaft,  but  no 
energy  is  lost  if  the  material  is  not  stressed  beyond  the  shearing 
elastic  limit.  Here  the  angle  of  twist  for  a  round  shaft  varies 
directly  as  the  length  of  the  shaft  and  also  directly  as  the  trans- 
mitted power,  but  it  varies  inversely  as  the  square  of  the  section 
area.  Some  theories  of  electricity  and  magnetism  appear  to 
indicate  that  forces  of  shearing  and  torsion  in  the  ether  may  in 
large  part  account  for  the  observed  phenomena.  Reiff's  Elas- 
ticitat  und  Elektricitat  (Leipzig,  1893)  contains  a  theory  of 
electricity  developed  from  the  fundamental  equations  of  the 
mathematical  theory  of  elasticity. 

ART.  193.     MISCELLANEOUS  PROBLEMS 

Below  are  given  a  number  of  topics  which  have  not  been 
treated  in  the  preceding  pages,  as  their  discussion  properly  be- 
longs to  special  works  on  special  branches  of  applied  mechanics. 


492  APPENDIX  AND  TABLES 

Teachers  who  wish  to  give  prize  problems  to  their  classes  may 
perhaps  find  some  of  these  useful  for  that  purpose. 

Prob.  193#.  Discuss  a  screw  with  square  threads  at  the  end  of  a 
bolt,  and  find  its  length  in  order  that  its  shearing  strength  may  be 
equal  to  the  tensile  strength  of  the  bolt. 

Prob.  193$.  An  I  beam  has  the  depth  d,  width  x,  thickness  of 
flange  b\t  and  thickness  of  web  b<%.  Find  the  value  of  x  so  that  the 
moment  of  inertia  about  an  axis  through  the  middle  of  the  web  shall 
be  equal  to  the  moment  of  inertia  about  an  axis  through  the  centei 
of  gravity  and  normal  to  the  web. 

Prob.  193r.  When  an  I  section  has  the  proportions  required  in 
the  last  problem,  show  that  all  moments  of  inertia  about  an  axis 
through  the  center  of  gravity  are  equal. 

Prob.  193^/.  A  load  Pis  supported  by  three  strings  of  equal  size 
hung  in  the  same  vertical  plane  from  the  ceiling  of  a  room.  The  mid- 
dle string  is  vertical  and  each  of  the  others  makes  an  angle  0  with  it. 
If  PI  is  the  stress  on  the  middle  string  and  P2  the  stress  on  each  of  the 
otheis,  show  that, 

Pl  =  P/(i  +  2  COS3fl)  P2  =  P  COS2#/(l  +  2  COS30) 

To  solve  this  problem  the  condition  must  be  introduced  that  the  inter- 
nal work  of  all  the  stresses  is  a  minimum. 

Prob.  193^.  A  load  P  is  supported  by  three  strings  of  equal  size 
lying  in  the  same  plane.  The  middle  string  is  vertical,  one  string 
makes  with  it  the  angle  0  on  one  side,  and  the  second  string  makes 
with  it  the  angle  0  on  the  other  side.  Find  the  stresses  in  the  strings. 

Prob.  193/.  A  circular  ring  of  mean  diameter  d  is  pulled  in  the 
direction  of  a  diameter  by  two  tensile  forces  each  equal  to  P.  Show 
that  the  maximum  bending  moment  is  at  the  section  where  P  is  ap- 
plied, and  that  its  value  is  %Pd3/7z(d2  +  4r2),  where  r  is  the  radius  of 
gyration  of  the  cross-section  of  the  ring. 

Prob.  193^.  An  elliptical  chain  link  has  the  mean  length  4^1  and 
the  mean  width  2^1,  where  d±  is  the  diameter  of  the  round  section 
area.  When  an  open  chain  link  of  these  proportions  is  subject  to  the 
tension  P  in  the  direction  of  its  length,  the  greatest  bending  moment 
occurs  where  the  tension  is  applied,  and  the  greatest  unit-stress  is 
S.iP/di2.  For  a  chain  link  with  a  cross  stud,  the  greatest  unit-stress 
is  z.qP/d2.  These  results  are  correct  only  when  the  elastic  limit  of 
the  material  is  not  exceeded. 


ART.  194  ANSWERS   TO  PROBLEMS  493 


ART.  194.    ANSWERS  TO  PROBLEMS 

The  following  are  the  answers  to  some  of  the  problems  given 
in  the  preceding  pages,  the  number  of  the  problem  being  in 
parenthesis.  The  instructions  in  Art.  8  should  be  carefully 
followed  by  the  student,  and  in  no  event  should  he  refer  to  an 
answer  until  the  solution  of  the  problem  is  completed.  How- 
ever satisfactory  it  may  be  for  a  student  to  be  able  to  know  that 
his  solutions  give  the  correct  results,  it  is  well  for  him  to  keep 
in  mind  that  in  actual  engineering  work  the  solutions  of  problems 
will  never  be  given. 

(16)  4.16  inches.  (26)  3.33  inches.  (30)  55  400  pounds  per  square 
inch.  (56)  i  780  feet.  (7a)  2.71  and  3.33  inches.  (86)  3.57  centi- 
meters. (90)  26300000  pounds  per  square  inch.  (146)  122  foot- 
pounds. (15)  i'  09".  (17)  26.2  square  inches.  (18a)  183.5  kilo- 
grams. (21)  $1058.84.  (23)  3  140  pounds.  (250)  4.04  cents.  (26a) 
105  ooo  pounds.  (29a)  Pi/P2=3/2.  (300)  0.88  inches.  (310)  2  500 
pounds  per  square  inch.  (32)  28  ooo  pounds  per  square  inch.  (346) 
3^  inches;  72  percent.  (350)  2\  inches.  (39)  ^=242,  F=i4o 
pounds.  (41)  960  pounds  per  square  inch.  (426)  c= ^(26+ 6')/(6+6'). 
(450)  8£  miles.  (496)  6000  pounds.  (500)  5;  2.5.  (506)  6  feet  5 
inches.  (51)  5  610  and  3  170  pounds  per  square  inch.  (520)  0.018 
inches.  (540)  14  500  ooo  pounds  per  square  inch.  (556)  74  500  kilo- 
grams per  square  centimeter.  (560)  8  to  3 ;  64  to  9.  (586)  0.72  inches. 
(596)  RI  =  290  pounds.  (606)  K=  0.3 66;  £=0.577.  (610)  l/m=  2.828. 
(62a)  Kl/(i  +  2K)  and  (2- K)l/($-2K).  (636)  0.027  inches.  (66c) 
o°25'47".  (71a)  —  jf^wl.  (73)  #1  =  —  113  pounds.  (740)  0.0131  wP/EI. 
(750)  ^=0.6095.  (766)  /i- 268.3,  ^2= 441-5  inches4.  (770)  r= 
I(dl2  +  d22)*.  (790)  5.05  inches.  (810)  3$.  (820)  23000  pounds. 
(S3)  13$  and  16*  inches.  (890)  30  pounds.  (896)  105  degrees.  (906) 
691  pounds  per  square  inch.  (916)  i.oi  horse-powers.  (920)  10.7. 
(926)  144  to  100.  (936)  9  380  ooo  pounds  per  square  inch.  (966)  2^ 
inches.  (996)  100  to  79.  (1010)  4i  inches.  (1016)  Nearly  8  inches. 
(105)  Sn=9  420  pounds  per  square  inch  and  0=54°  13'-  (1070)  4.9. 
(108)  5^  =  205  pounds  per  square  inch.  (1190)  122  foot-pounds. 
(1200)  5j  horse-power.  (1216)  0.52  horse-powers.  (123)  3PP/2$6EI. 
(126)  —  \P.  (130)  1 20  feet  per  second.  (148)  2  886  ooo  pounds.  (149) 
S=  + 14  ooo  pounds  per  square  inch.  (150)  18  ooo  pounds  per  square 


494  APPENDIX  AND  TABLES 

inch.  (151)  54000  pounds  per  square  inch.  (1550)  Deduce  an  ex- 
pression for  RI  in  terms  of  the  given  radii  and  Se',  then  find  the  value 
of  r2  which  renders  RI  a  maximum.  (156)  6  rollers.  (157)  64  and  7. 
(1600)  r-(r2-$d2)*.  (163a)  1.6  inches.  (165)  About  1.4  inches.  (175) 
56°  19'  with  axis  of  bolt.  (177)  See  theory  of  equations  in  algebra. 
(180)  54°  44'  with  the  greater  apparent  stress. 


Evolvi  varia  problemata.  In  scientiis  enim  ediscendis  prosunt  ex- 
empla  magis  quam  praecepta.  Qua  de  causa  in  his  fusius  expatiatus 
sum.— NEWTON. 

Nous  avons  pour  but,  non  de  donner  un  traite  complet,  mais  de 
montrer,  par  des  exemples  simples  et  varies,  1'utilite  et  1'importance  de 
la  theorie  mathematique  de  1'elasticite. — LAME. 

Homo,  naturae  minister  et  interpres,  tantum  facit  et  intelligit  quan- 
tum de  naturae  ordine  re  vel  mente  observaverit,  nee  amplius  scit  aut 
potest. — BACON. 


ART.  195.     EXPLANATION  OF  TABLES 

Tables  1-5  give  average  physical  constants  for  materials 
and  Tables  6-13  give  properties  of  beams  and  columns.  At  the 
foot  of  each  table  is  a  reference  to  the  articles  where  its  use  is 
explained. 

Table  14  gives  weights  per  linear  foot  of  wrought-iron  bars 
both  square  and  round,  the  side  of  the  square  or  the  diameter 
of  the  circle  ranging  from  &  to  ipf  inches.  Approximate,  weights 
of  bars  of  other  materials  may  be  derived  from  this  table  by  the 
following  rules : 

for  Timber,  divide  by  12 

for  Brick,  divide  by  4 

for  Stone,  divide  by  3 

for  Cast  Iron,  subtract  6  percent 

for  Steel,  add  2  percent 

For  example,  a  cast-iron  bar  6J  inches  square  and  8  feet  long 
weighs  8(157.6  — 0.06X157.6)  =i  185  pounds.  In  like  manner 
a  steel  bar  2&  inches  in  diameter  and  4  feet  9  inches  in  length 
weighs  41(12.53  +  0.02X12.53)  =60.7  pounds. 


ART.  195  EXPLANATION    OF   TABLES  495 

Table  15  gives  four-place  squares  of  numbers  from  i.oo  to 
9.99,  the  arrangement  being  the  same  as  that  of  a  logarithmic 
table.  By  properly  moving  the  decimal  points,  four-place  squares 
of  other  numbers  may  also  be  taken  out.  For  example,  the 
square  of  0.874  is  0.7639,  that  of  87.4  is  7  639,  and  that  of  874 
is  763  900,  all  correct  to  four  significant  figures. 

Table  16  gives  four- place  areas  of  circles  for  diameters  rang- 
ing from  i.oo  to  9.99,  arranged  in  the  same  manner.  By  properly 
moving  the  decimal  point,  four-place  areas  for  all  circles  may 
be  found.  For  instance,  if  the  diameter  is  4.175  inches,  the 
area  is  13.69  square  inches;  if  the  diameter  is  0.535  inches,  the 
area  is  0.2248  square  inches;  if  the  diameter  is  12.2  feet,  the 
area  is  116.9  square  feet,  all  correct  to  four  significant  figures. 

Table  17  gives  four- place  trigonometric  functions  of  angles, 
and  Table  18  the  logarithms  of  these  functions.  The  term 
'arc'  means  the  length  of  a  circular  arc  of  radius  unity,  or  the 
value  of  the  angle  in  radians,  while  'coarc'  is  the  complement 
of  the  arc.  If  6  is  the  number  of  degrees  in  an  angle,  the  value 
of  the  arc  is  nd/i8o,  and  this  subtracted  from  \K  gives  the  coarc. 

Table  19  gives  four-place  logarithms  of  numbers  and  these 
are  sufficiently  precise  for  nearly  all  computations  arising  iu 
the  application  of  the  principles  of  mechanics  of  materials.  Thft 
differences  in  the  last  column  enable  interpolations  to  be  made 
so  that  four-place  logarithms  of  numbers  with  four  significant 
figures  may  be  taken  out;  for  example,  the  logarithm  of  0.6534 
is  1.8152. 

Table  20,  taken  from  the  author's  Elements  of  Precise  Sur- 
veying and  Geodesy,  gives  mathematical  constants  and  their 
logarithms  to  nine  decimals ;  this  is  a  greater  number  of  decimals 
than  will  ever  be  needed  in  computations  on  the  materials  of 
engineering,  but  they  will  sometimes  be  required  for  the  dis- 
cussion of  geodetic  and  physical  measurements. 


496 


TABLES 


TABLE  1.    AVERAGE  WEIGHT  AND  EXPANSIBILITY 


Material 

Weight  per  Unit  Volume 

Coefficient  of  Linear 
Expansion 

Pounds  per 
cubic  foot 

Kilograms 
per  cubic 
meter 

For  1° 

Fahrenheit 

For  1° 
Centigrade 

Brick 

125 

2  OOO 

o.ooo  0050 

o.ooo  0090 

Concrete 

ISO 

2  400 

o.ooo  0055 

o.ooo  0099 

Stone 

1  60 

2  560 

o.ooo  0050 

o.ooo  0090 

Timber 

40 

600 

0.000  0020 

o.ooo  0036 

Cast  Iron 

450 

7  200 

o.ooo  0062 

0.000  OII2 

Wrought  Iron 

480 

7700 

o.ooo  0067 

O.OOO  01  2  I 

Structural  Steel 

49° 

7800- 

o.ooo  0065 

o.ooo  0117 

Strong  Steel 

49  1 

7800 

o.ooo  0065 

o.ooo  0117 

Explanation  in  Arts.  17  and  100 


TABLE  2.    AVERAGE  ELASTIC  PROPERTIES 


Elastic  Limit 

Modulus  of  Elasticity 

Material 

Pounds  per 
square  inch 

Kilograms 
per  square 
centimeter 

Pounds  per 
square  inch 

Kilograms 
per  square 
centimeter 

Brick 

I  OOO 

70 

2  OOO  OOO 

140  ooo 

Concrete 

800 

56 

2  500  ooo 

175  ooo 

Stone 

2  OOO 

140 

6  ooo  ooo 

420  ooo 

Timber 

3  ooo 

210 

i  500  ooo 

105  ooo 

Cast  Iron             { 

6  ooo 

420 

15  ooo  ooo 

i  050  ooo 

20  000 

I  400 

15  ooo  ooo 

i  050  ooo 

Wrought  Iron 

25  ooc 

I   750 

25  ooo  ooo 

i  750  ooo 

Structural  Steel 

35000 

2450 

30  ooo  ooo 

2  IOO  OOO 

Strong  Steel 

50  ooo 

35°° 

30  ooo  ooo 

2  IOO  000 

Explanation  in  Arts.  2  and  9.  Values  for  Brick,  Concrete,  and  Stone  are 
for  compression  only.  For  Cast  Iron  the  upper  values  apply  to  tension  and  the 
lower  ones  to  compression.  For  other  materials  the  values  apply  to  both 
tension  and  compression. 


TABLES 


497 


TABLE  3.    AVERAGE  TENSILE  AND  COMPRESSIVE  STRENGTH 


Ultimate  Tensile  Strength 

Ultimate  Compressive 
Strength 

Material 

Pounds  per 
square  inch 

Kilograms 
per  square 
centimeter 

Pounds  per 
square  inch 

Kilograms 
per  square 
centimeter 

Brick 

3000 

2IO 

Concrete 

300 

21 

3000 

210 

Stone 

6  ooo 

420 

Timber 

IO  OOO 

700 

8  ooo 

560 

Cast  Iron 

20  ooo 

I  400 

90  ooo 

6300 

Wrought  Iron 

50  ooo 

35°° 

50  ooo 

35°° 

Structural  Steel 

60  ooo 

4  200 

60  ooo 

4  200 

Strong  Steel 

IOO  000 

7  ooo 

1  2O  OOO 

8  400 

Explanation  in  Arts.  4,5,  19-25 


TABLE  4.  AVERAGE  SHEARING  AND  FLEXURAL  STRENGTH 


Material 

Ultimate  Shearing  Strength 

Modulus  of  Rupture. 

Pounds  per 
square  inch 

Kilograms 
per  square 
centimeter 

Pounds  per 
square  inch 

Kilograms 
per  square 
centimeter 

Brick 

700 

5° 

800 

55 

Concrete 

I  2OO 

84 

Stone 

I  500 

I°5 

2  OOO 

140 

Timber,  with  grain 

500 

35 

Timber,  across  grain 

3000 

210 

9  ooo 

630 

Cast  Iron 

20  000 

I  400 

35000 

2  40O 

Wrought  Iron 

40  ooo 

2  800 

Structural  Steel 

50  ooo 

35°° 

Strong  Steel 

75000 

5  20° 

no  ooo 

7700 

Explanation  in  Arts.  6   19-25,  52 


498 


TABLES 


TABLE  5.    WORKING  UNIT-STRESSES  FOR  BUILDINGS 

ABSTRACTED    FROM    THE    BUILDING   CODE   OF   THE    CITY   OF   NEW   YORK,    N.  Y. 


Material 

") 
Pounds  per  Square  Inch 

Tension 

Com- 
pression 

Shear 

Flexure 

Hemlock 
Spruce 
White  Pine 
Yellow  Pine 
Oak 

600 
800 
800 
I  20O 
I  000 

500 
800 
800 

I  OOO 

900 

275* 
320* 
250  * 
500  * 
600* 

600 
800 
800 
I  200 

I  OOO 

Brick 
Brickwork 

300 

5° 

In  Portland  Cement  Mortar 
In  Natural  Cement  Mortar 
In  Lime  Mortar 

250 
208 

III 

30 
3° 

Concrete 

Portland  Cement,  i,  2,  4 
Natural  Cement,  i,  2,  4 

230 

I25 

30 
16 

Rubble  Stonework 

In  Portland  Cement  Mortar 
In  Natural  Cement  Mortar 

140 
in 

In  Lime  Mortar 

70 

4 

Sandstone 

I  000 

IOO 

Limestone 
Marble 
Granite 
Slate 

I  500 
900 

I  700 

I  OOO 

15° 

120 

180 
400 

Cast  Iron 

Wrought  Iron 
Wrought-iron  shop  rivets 
Wrought-iron  field  rivets 

3000 

12  OOO 

1  6  ooo 

12  000 

3000  1 

6  ooo 
7500 
6  ooo 

3  ooo  Ten. 
16000  Comp. 

12  OOO 

Rolled  Steel 

1  6  ooo 

1  6  ooo 

9  ooo 

16000 

Steel  shop  rivets 
Steel  field  rivets 

10  000 

8  ooo 

Cast  Steel 

1  6  ooo 

1  6  ooo 

• 

Explanation  in  Art.  7.          *  Across  grain. 


TABLES 


499 


TABLE  6.    STEEL  I-BEAM  SECTIONS 


Depth 

of 
Beam 

Inches 

Weight 
Foot 

Pounds 

Width 
of 
Flange 

Inches 

Section 
Area 
a 

Sq.  In. 

Axis  perpendicular  to  web 

Axis  parallel  to  web 

Moment 
Inertia 
/ 
Inches4 

Section 
Factor 
I/e 
Inches3 

Radius 
Gyration 
r 
Inches 

Moment 
Inertia 
/ 
Inches4 

Radius 
Gyration 
r 
Inches 

24 

100 

7.25 

29.41 

2380 

198.4 

9.00 

48.56 

1.28 

*24 

80 

7.00 

23-32 

2088 

174.0 

9.46 

42.86 

I.36 

20 

100 

7.28 

29.41 

1656 

165.6 

7-50 

52-65 

1-34 

*20 

80 

7.00 

23-73 

1467 

146.7 

7.86 

45.81 

1-39 

*20 

65 

6.25 

19.08 

1170 

117.0 

7.83 

27.86 

I.  21 

18 

70 

6.26 

20.59 

921.3 

102.4 

6.69 

24.62 

1.09 

*i8 

55 

6.00 

15-93 

795-6 

88.4 

7.07 

21.  19 

1-15 

15 

100 

6.77 

29.41 

900.5 

120.  I 

5-53 

50.98 

1-31 

•15 

80 

6.40 

23.81 

795-5 

106.  i 

5-78 

41.76 

1.32 

*iS 

60 

6.00 

17.67 

609.0 

81.2 

5-87 

25.96 

I  .21 

•is 

42 

5-5° 

12.48 

441.7 

58-9 

5-95 

14.62 

1.  08 

12 

55 

5-6i 

16.18 

321.0 

53-5 

4-45 

17.46 

1.04 

*I2 

40 

5-25 

11.84 

268.9 

44.8 

4-77 

I3.8I 

1.  08 

*I2 

3*1 

5.00 

9.26 

215.8 

36.0 

4-83 

9-50 

1.  01 

10 

40 

5.10 

11.76 

158-7 

3i-7 

3-67 

9-5° 

0.90 

*IO 

25 

>4.66 

7-37 

122.  I 

24.4 

4.07 

6.89 

0.97 

9 

35 

'4-77 

10.29 

III.  8 

24.8 

3-29 

7-31 

0.84 

*9 

21 

4-33 

6.31 

84.9 

18.9 

3-67 

5-i6 

0.90 

8 

25i 

4.27 

7-5° 

68.4 

17.1 

3.02 

4-75 

0.80 

*  8 

18 

4.00 

5-33 

56-9 

14.2 

3-27 

3-78 

0.84 

7 

20 

3-87 

5.88 

42.2 

12.  I 

2.68 

3-24 

0.74 

*  7 

15 

3.66 

4.42 

36.  2 

10.4 

2.86 

2.67 

0.78 

6 

17} 

3-57 

5-07 

26.2 

8.7 

2.27 

2-36 

0.68 

*  6 

I2i 

3-33 

3.61 

21.8 

7-3 

2.46 

1.85 

0.72 

5 

14* 

3-29 

4-34 

IS-2 

6.1 

1.87 

1.70 

0.63 

*  - 

D 

9* 

3.00 

2.87 

12.  I 

4.8 

2.05 

1.23 

0.65 

4 

10* 

2.88 

3-°9 

7-i 

3-6 

i-52 

i  .01 

0.57 

*  4 

7* 

2.66 

2.21 

6.0 

3-° 

1.64 

0.77 

o-59 

3 

7i 

2.52 

2.21 

2-9 

1.9 

i.  IS 

0.66 

0.52 

*  3 

Si 

2-33 

1.63 

2-5 

i-7 

1.23 

0.46 

o-53 

Explanation  in  Arts.  44  and  51.         *  Standard  sizes,  others  are  special. 


500 


TABLES 


TABLE  7.     STEEL  BULB-BEAM  SECTIONS 


Depth 
Inches 

Weight 
Foot 

Pounds 

Section 
Area 

Sq.  In. 

Axis  perpendicular  to  web 

1 
Axis  parallel  to  web| 

Moment 
Inertia 
/ 
Inches4 

Base  to 
Neutral 
Axis 
Inches 

Section 
Factor 
I/c 

Inches3 

Radius 

Gyration 
r 
Inches 

Moment 
Inertia 
/ 
Inches4 

Radius 
Gyration 
r 
Inches 

II* 

32.2 

9-51 

179-3 

5-°7 

27.9 

4-34 

6.36 

C.82 

10 

28.0 

8.20 

118.5 

4.28 

20.7 

3.80 

6.08 

0.86 

9 

25.0 

7-35 

85.0 

3-90 

l6.7 

3-40 

4.85 

0.81 

8 

21  .0 

6.17 

57-8 

3-48 

12.8 

3.06 

3.58 

0.76 

7 

18.0 

S-32 

37-0 

3-°4 

9-3 

2.64 

2.56 

0.69 

6 

14.5 

4.27 

21.8 

2.61 

6.4 

2.26 

1.62 

0.62 

5 

11  -5 

3-39 

12.0 

2.22 

4-3 

1.88 

I  .01 

o-55 

Explanation  in  Art.  44 


TABLE  8.     STEEL  T  SECTIONS 


Axis  perpendicular  to  web 

Axis  parallel  to  web 

Size 

WirH-Vl 

Weight 

Section 

w  idtn 

V>v 

per 

Area 

oy 
Depth 

Foot 

a 

Moment 
Inertia 

Base  to 

Neutral 

Section 
Factor 

Radius 
Gyration 

Momsnt 
Inertia 

Radius 
Gyration 

1 

Axis 

/  c 

r 

/ 

r 

Inches 

Pounds 

Sq.  In. 

Inches4 

Inches 

Inches3 

Inches 

Inches4 

Inches 

6X4 

17.4 

5-12 

6.56 

1  .00 

2.19 

I.I3 

9-33 

1-35 

5X4 

15-3 

4-54 

6.16 

1.  08 

2.  II 

I.I7 

5-41 

1.09 

4X4 

10.9 

3.10 

4.70 

I-I5 

1.64 

1.23 

2.20 

0.85 

4X3 

9.0 

2.67 

1.99 

0.78 

0.90 

0.87 

2.  10 

0.89 

3*X3i 

7.0 

2.08 

2.27 

0.94 

0.89 

1.04 

1.03 

0.94 

3iX3 

7-o 

2.  II 

I.65 

0.80 

0-75 

0.88 

1.18 

0-75 

3X3 

6-5 

I.9I 

i-57 

0.87 

0.74 

0.91 

0-75 

0.62 

3X2} 

5-o 

1.46 

0.78 

0.66 

0.42 

0-73 

0.60 

0.64 

2}X3 

6.0 

I.76 

1.48 

o-93 

0.71 

0.92 

0.44 

0.50 

2iX2$ 

5-8 

I.7I 

°-95 

0.76 

°-55 

°-75 

0.48 

0-53 

2X2 

3-5 

1.03 

Q-37 

0.60 

0.26 

0.60 

0.18 

0.41 

2Xl£ 

3-o 

0.91 

0.16 

o-45 

0.15  ' 

0.42 

0.17 

0.44 

2X1 

2-5 

0.72 

0.05 

0.27 

0.07 

0.26 

0.17 

0.49 

Explanation  in  Art.  44 


TABLES 


501 


TABLE  9.     STEEL  CHANNEL  SECTIONS 


Depth 
Inches 

Weight 

Foot 

Pounds 

Width 
of 
Flange 

Inches 

Section 
Area 
a 

Sq.  In. 

Axis  perpendicular 
to  web 

Axis  parallel  to  web 

Moment 
Inertia 

I 

Inches4 

Radius 
Gyration 
r 

Inches 

Moment 
Inertia 
/ 

Inches4 

Radius 
Gyration 
r 

Inches 

Outside 
of  Web  to 
Center  of 
Gravity 

Inches 

15 

55 

3.82 

16.18 

430  .  2 

5-l6 

I3-I9 

0.87 

0.82 

15 

45 

3.62 

13.24 

375  -1 

5-32 

10.29 

0.88 

0.79 

15 

35 

3-43 

10.29 

320.0 

5-58 

8.48 

0.91 

0.79 

*i5 

33 

3-40 

9.90 

312.6 

5.62 

8.23 

0.91 

0.79 

12 

40 

3-42 

11.76 

197.0 

4.09 

6.63 

Q-75 

0.72 

12 

3° 

3-i7 

8.82 

161  .7 

2.28 

5-2i 

0.77 

0.68 

12 

25 

3-°5 

7-35 

144.0 

4-43 

4-53 

0.78 

0.68 

*I2 

2oi 

2.94 

6.03 

128.1 

4.61 

3-9i 

0.80 

0.68 

10 

35 

3-i8 

10.29 

ii5-5 

3-35 

4.66 

0.67 

0.70 

10 

25 

2.89 

7-35 

91  .0 

3-52 

3-4o 

0.68 

0.62 

10 

20 

2.74 

5.88 

78.7 

3.66 

2.85 

0.70 

0.61 

*IO 

15 

2.60 

4.46 

66.9 

3-87 

2.30 

0.72 

0.64 

9 

25 

2.81 

7-35 

70.7 

'3.10 

2.98 

0.64 

0.61 

9 

J5 

2.49 

4.41 

5°-9 

3-40 

i-95 

0.66 

o-59 

*9 

13* 

2-43 

3-89 

47-3 

3-49 

1.77 

0.67 

0.60 

8 

2li 

2.62 

6.25 

47-8 

2-77 

2.25 

0.60 

0.58 

8 

i6J 

2.44 

4.78 

39-9 

2.89 

1.78 

0.61 

0.56 

*  8 

"i 

2  .  26 

3-35 

32-3 

3-n 

i-33 

0.63 

0.58 

7 

i9f 

2-5r 

5-8i 

33-2 

2-39 

1-85 

0.56 

0.58 

7 

I4f 

2.30 

4-34 

27.2 

2.50 

1.40 

o-57 

0-54 

*  7 

9f 

2.09 

2.85 

21.  I 

2.72 

0.98 

o-59 

o-55 

6 

rSi 

2.28 

4-56 

19-5 

2.07 

1.28 

o-53 

o-55 

6 

loj 

2.04 

3-09 

I5-I 

2  .  21 

0.88 

o-53 

0.50 

*  6 

8 

1.92 

2.38 

I3.0 

2-34 

0.70 

0.54 

0.49 

5 

iii 

2.04 

3.38 

10.4 

1-75 

0.82 

0.49 

0.51 

*  5 

6* 

i-75 

i-95 

7-4 

i-95 

0.48 

0.50 

0.49 

4 

7i 

1.72 

2.13 

4.6 

i  .46 

0.44 

0-45 

0.46 

*  4 

5* 

1.58 

i-55 

3-8 

1.56 

0.32 

0-45 

0.46 

3 

6 

i.  60 

1.76 

2.1 

i.  08 

0.31 

0.42 

0.46 

*  3 

4 

1.41 

1.19 

1.6 

1.17 

0.20 

0.41 

0.44 

Explanation  in  Arts.  44  and  76.         *  Standard  sizes:   others  are  special. 


502 


TABLES 


TABLE  10.     STEEL  ANGLE  SECTIONS 


Size 
of  Angle 

Inches 

Weight 
Foot 

Pounds 

Section 
Area 

Sq.  In. 

Axis  parallel  to  long  leg 

Axis  parallel  to  short  leg 

Axis 
to  back 
of  leg 
Inches 

Moment 
Inertia 

Inches4 

Radius 
Gyra- 
tion 
Inches 

Axis 
to  back 
of  leg 
Inches 

Moment 
Inertia 

Inches4 

Radius 
Gyration 

Inches 

6X4X1 

30.6 

9.00 

I.I7 

10-75 

1.09 

2.17 

30.75 

1.85 

6X4X1 

23-6 

6.94 

1.  08 

8.68 

I  .12 

2.08 

24.51 

1.88 

6X4Xi 

16.2 

4-75 

0.99 

6.27 

I-I5 

1.99 

17.40 

1.91 

5X4X1 

24.2 

7.II 

I.  21 

9-23 

1.14 

1.71 

16.42 

1.52 

5X4X| 

21  .  I 

6.19 

1.16 

8.23 

I-IS 

1.66 

14.60 

1-54 

5X4Xi 

14-5 

4.25 

1.07 

5-96 

1.18 

1-57 

10.46 

i-57 

5X3Xf 

19.9 

5.84 

0.86 

3-7i 

0.80 

1.86 

13.98 

i-55 

5X3X£ 

12.8 

3-75 

0-75 

2.58 

0.83 

i-75 

9-45 

i-59 

4X3Xf 

16.0 

4.69 

0.92 

3.28 

o».84 

1.42 

6-93 

I  .22 

4X3X| 

ii.  i 

3-25 

0.83 

2.42 

0.86 

1-33 

5-°5 

1.25 

3*X3Xi 

10.2 

3-°o 

0.88 

2-33 

0.88 

1-13 

3-45 

1.07 

3iX2|X^ 

9.4 

2-75 

o.  70 

1-36 

0.70 

1.20 

3-24 

1.09 

3XajXJ 

8.5 

2.50 

0-75 

1.30 

0.72 

I.OO 

2.08 

0.91 

3X2X^ 

7-7 

2.25 

0.58 

0.67 

0-55 

1.  08 

1.92 

0.92 

Equal 

legs 

Axis 

parallel 

to  leg 

Least  r 

6X6X1 

37-4 

II.OO 

1.86 

35-46 

1.  80 

1.16 

6X6Xf 

28.7 

8.44 

1.78 

28.15 

1.83 

1.17 

6X6Xi 

19.6 

5-75 

1.68 

19.91 

1.86 

«.  2*^60 

glaJ[ 

I.l8 

5X5X1 

30.6 

9.00 

1.61 

19.64 

1.48 

?  jj?  8 

0.96 

5X5X1 

23.6 

6.94 

1.52 

15-74 

1.51 

0.97 

5X5Xi 

16.2 

4.75 

1-43 

11.25 

1-54 

2^1 

0.98 

4X4Xf 

18.5 

5-84 

1.29 

8.14 

1.18 

111! 

0.77 

4X4Xi 

12.8 

3-75 

1.18 

5.56 

1.22 

U5  O   <U  \r> 

8  «  «  * 

0.78 

3iX3iXi 

II  .  I 

3-25 

i.  06 

3-64 

1.  06 

ajjfl 

0.68 

3X3Xi 

9-4 

2-75 

0.93 

2.22 

0.90 

ill 

0.58 

2jX2^X^ 

7-7 

2.25 

0.81 

1.23 

0.74 

0.47 

2X2X1 

3-2 

0.94 

o-59 

0-35 

0.61 

o-39 

Explanation  in  Art.  44 


TABLES 


503 


TABLE  11.    STEEL  Z  SECTIONS 


Size 

Weight 
Foot 

Thick- 
ness 

Section 
Area 

Moment 
Inertia 
la 

Moment 
Inertia 
Ib 

Tangent 
of  Angle 
between 

Least 
Radius 
Gyration 

Inches 

Inches 

Sq.  In. 

Inches4 

Inches4 

/2  and  /£ 

Inches 

8X3 

16.9 

1 

4-97 

44.64 

5.6o 

0.27 

0.72 

7iX3 

l6.3 

I 

4.78 

38.19 

5-59 

0.29 

0.72 

6X3i 

29-3 

f 

8.63 

42.12 

15-44 

0.52 

0.81 

5X3i 

23-7 

ft 

6.96 

23.68 

n-37 

0.62 

o-73 

4X3^ 

18.9 

1 

5-55 

12.  II 

8-73 

0.81 

0.65 

3X2^ 

I2.5 

i 

3-69 

4-59 

4-85 

0.965 

o.53 

Explanation  in  Arts.  44  and  166 


TABLE  12.    COMPARISON  OF  BEAMS 


Beams  of  Uniform  Cross-section 

Maximum 
Moment 

Maximum 
Deflection 

Relative 
Strength 

Relative 
Stiffness 

Cantilever,  single  load  at  end 

Wl 

i  Wl3 
3    El 

I 

I 

Cantilever,  uniform  load 

Vn 

i  Wl3 
8  ~EI 

2 

0 

*3 

Simple  beam,  load  at  middle 

\Wl 

i  Wl3 

4 

16 

Simple  beam  uniformly  loaded 

ft* 

j_WP 

8 

«t 

Beam  fixed  at  one  end,  supported 
at  other,  load  near  middle 

O.IC)2Wl 

Wl3 
0.0008  — 

S-2 

30-7 

Beam  fixed  at  one  end,  supported 
at  other,  uniform  load 

Wl 

Wl3 
0.0054  _ 

8 

62 

Beam  fixed  at  both  ends,  load  at 
middle 

\Wl 

i    Wl3 
192    El 

8 

64 

Beam  fixed  at  both  ends,  uniform 
load 

\Wl 

i    Wl3 
~&4~EI 

12 

128 
i 

Explanation  in  Arts.  56  and  63 


504 


TABLES 


TABLE  13.  GERMAN  I  BEAMS 


Axis  perpendicular  to  web 

Axis  parallel  to  web 

Depth 

Weight 

Width 

Section 

of 
Beam 

Meter 

of 
Flange 

Area 
a 

Moment 

Section 

Radius 

Moment 

Radius 

Inertia 

Factor 

Gyration 

Inertia 

Gyration 

/ 

I/c 

r 

/ 

r 

cm. 

kilos 

cm. 

sq.  cm. 

cm.4 

cm.3 

cm. 

cm.4 

cm. 

5° 

J59-37 

19.0 

204.32 

78  040 

3  I22 

J9-54 

3060 

3.87 

45 

129.  ii 

17.1 

l65-52 

51  230 

2   277 

*7-59 

2  OO5 

3.48 

40 

103.54 

15   6 

132.74 

32  680 

I   634 

15.69 

1358 

3-20 

35 

80.78 

14.1 

!03-57 

19  690 

I    125 

T3-79 

871.0 

2.90 

32 

68.63 

13.2 

87.99 

13970 

873.3 

12.60 

651.0 

2.72 

3° 

60.82 

12.6 

77-97 

II  000 

733-1 

11.88 

539-3 

2.63 

28 

61.50 

15.0 

78.85 

10  280 

733-9 

ii  .41 

827.7 

3-24 

28 

53-55 

12.0 

68.65 

8523 

608.8 

ii  .  14 

439-4 

2-53 

26 

46.89 

ri.  4 

60.  ii 

6413 

493-3 

10  -33 

343-3 

2-39 

25 

43-66 

ii  .  i 

55-97 

5554 

444-3 

9-96 

306.6 

2-34 

24 

46.64 

J3-5 

59-So 

5  776 

481.3 

9-83 

516.9 

2.94 

24 

40.54 

10.8 

51-97 

4784 

399-7 

9-59 

272.5 

2.29 

23 

37-56 

10.5 

48.15 

4097 

356.3 

9.22 

241.6 

2.24 

22 

41.42 

J3-5 

53  -to 

4347 

395-2 

9-°5 

459-0 

2-94 

22 

34-73 

IO  .  2 

44-52 

3434 

312.2 

8.78 

205.8 

2-15 

21 

31.96 

9.9 

40.98 

2898 

276.0 

8.40 

180.7 

2.10 

2O 

29.29 

9.6 

37-55 

2  428 

242.8 

8.04 

T54-7 

2.03 

18 

32.06 

13-5 

41  .10 

2  363 

262.8 

7.58 

369-9 

3.00 

18 

24-34 

9.0 

31.20 

i  661 

184.6 

7-30 

119.8 

1.96 

16 

19.83 

8.4 

25-43 

i  067 

133-4 

6.48 

83-3 

1.81 

15 

17.60 

8.0 

22.57 

840.0 

112.  0 

6.  10 

68.3 

1.74 

14 

16.02 

7.6 

20.54 

659-4 

94-2 

5-67 

55-9 

1.65 

13 

14.56 

7.2 

18.67 

523-9 

80.6 

5-29 

47-8 

i.  60 

12 

12.69 

6.8 

16.27 

392.4 

65.4 

4.91 

37-6 

1.52 

10 

9.69 

6.0 

12.42 

208.0 

41  .6 

4.08 

22.6 

J-35 

8 

7.07 

5-2 

9.07 

97.2 

24-3 

3-17 

12.8 

1.19 

Explanation  in  Art.  44 


TABLES 


505 


TABLE  14.    WEIGHT  OF  WROUGHT-!RON  BARS 


Side  or 
Diam- 
eter 
Inches 

Pounds  per  Linear 
Foot 

Side  or 
Diam- 
eter 
Inches 

Poundsper  Linear 
Foot. 

Side  or 
Diam- 
eter 
Inches 

Poundsper  Linear 
Foot 

Square 
Bars 

Round 
Bars 

Square 
Bars 

Round 
Bars 

Square 
Bars 

Round 
Bars 

O 

2 

13.33 

10.47 

5 

83.33 

65.45 

A 

0.013 

O.OIO 

A 

14.18 

11.14 

i 

87.55 

68.76 

* 

O.O52 

0.041 

i 

15.05 

11.82 

i 

91.88 

72.16 

* 

O.H7 

0.092 

A 

15.95 

12.53 

1 

96.30 

75.64 

* 

O.208 

0.164 

i 

16.88 

13.25 

i 

100.8 

79.19 

A 

0.326 

0.256 

A 

17.83 

14.00 

i 

105.5 

82.83 

1 

0.469 

0.368 

1 

18.80 

14.77 

1 

110.  2 

86.56 

A 

0.638 

0.501 

A 

19.80 

15.55 

1 

II5.I 

90.36 

1 

0.833 

0.654 

1 

20.83 

16.36 

6 

120.0 

94.25 

A 

1.055 

0.828 

T°* 

21.89 

17.19 

i 

I25.I 

98.22 

I 

1.302 

1.023 

I 

22.97 

18.04 

i 

130.2 

102.3 

H 

1.576 

1.237 

H 

24.08 

18.91 

1 

135.5 

106.4 

* 

1.875 

1.473 

4 

25.21 

19.80 

1 

140.8 

no.  6 

if 

2.2OI 

1.728 

i! 

26.37 

20.71 

f 

146.3 

114.9 

i 

2.552 

2.004 

i 

27.55 

21.64 

f 

251.9 

119.3 

H 

2.930 

2.301 

it 

28.76 

22.59 

1 

157.6 

123.7 

i 

3-333 

2.618 

3 

30.00 

23.56 

7 

166.3 

128.3 

A 

3.763 

2.955 

i 

32.55 

25.57 

i 

175-2 

137-6 

i 

4.219 

3.313 

i 

35.21 

27.65 

1 

187.5 

147.3 

A 

4.701 

3.692 

1 

37.97 

29.82 

1 

2OO.2 

157.2 

i 

5.208 

4.091 

1 

40.83 

32.07 

8 

213.3 

167.6 

T<V 

5.742 

4.510 

I 

43.80 

34-40 

i 

226.9 

178.2 

1 

6.302 

4.950 

f 

46.88 

36.82 

i 

240.8 

189.2 

7 
T* 

6.888 

5.410 

* 

50.05 

39-31 

f 

255.2 

200.4 

i 

7.500 

5.890 

4 

53.33 

41.89 

9 

270.0 

212.  1 

A 

8.138 

6.392 

i 

56.72 

44-55 

i 

285.2 

224.O 

i 

8.802 

6.913 

i 

60.  21 

47.29 

1 

300.8 

236.3 

« 

9.492 

7-455 

$ 

63.80 

50.11 

f 

316.9 

248.9 

1 

10.21 

8.018 

1 

67.50 

53-01 

10 

333-3 

261.8 

if 

10.95 

8.601 

i 

71.30 

56.00 

i 

350.2 

275-1 

i 

11.72 

9.204 

i 

75-21 

59.07 

1 

367.5 

288.6 

H 

13    51 

9.828 

1 

79.22 

62.22 

f 

385.2 

302.5 

Explanation  in  Art.  188 


506 


TABLES 


TABLE  15.     SQUARES  or  NUMBERS 


n. 

01234 

56789 

Diff. 

I.O 

i.ooo    1.020    1.040    1.061    1.082 

1.103    1.124    i-T45    1-166    1.188 

22 

.[ 

1.210    1.232    1.254    1.277    I-3°° 

1.323    1.346    1.369    1.392    1.416 

24 

.2 

1.440    1.464    1.488    1.513    1.538 

1.563    1.588    1.613    1.638    1.664 

26 

•3 

1.690    1.716    1.742    1.769    1.796 

1.823    I-°SQ   1.877    1.904    1.932 

28 

4 

1.960    1.988    2.016   2.045    2.074 

2.IO3     2.132     2.l6l     2.190     2.22O 

3° 

•5 

2.250   2.280    2.310    2.341    2.372 

2.403     2.434     2.465     2.496     2.528 

32 

,6 

2.560   2.592    2.624   2.657    2.690 

2.723     2.756     2.789     2.822     2.856 

34 

•7 

2.890   2.924    2.958    2.993    3.028 

3.063     3.098     3.133     3.168     3.204 

36 

1.8 

3.240   3.276   3.312    3.349   3.386 

3.423     3.460     3.497     3.534     3.572 

38 

1.9 

3.610   3.648    3.686   3.725    3.764 

3.803     3.842     3.881     3.920     3.960 

40 

2.O 

4.000   4.040   4.080   4.121    4.162 

4.203     4.244     4.285     4.326    4.368 

42 

2.1 

4.410   4.452   4.494   4.537    4.580 

4.623     4.666    4.709    4.752     4.796 

44 

2.2 

4.840   4.884   4.928   4.973    5.018 

5.063     5.108     5.153     5.198     5.244 

46 

2-3 
2.4 

5.290    5.336    5.382    5.429    5.476 
5.760    5.808    5.856    5.905    5.954 

5-523     5-570     5.617     5.664     5.712 

6.003   6-°52   6.101    6.150  6.200 

48 
5° 

2.5 

6.250   6.300   6.350   6.401    6.452 

6.503   6.554   6.605   6.656   6.708 

52 

2.6 

6.760   6.812    6.864   6.917    6.970 

7.023   7.076   7.129   7.182   7.236 

54 

2.7 

•7.290   7.344   7.398   7.453   7.508 

7.563   7.618    7.673   7.728   7.784 

56 

2.8 

7.840   7.896  7.952   8.009   8.066 

8.123   8.180   8.237    8.294  8.352 

2.9 

8.410  8.468   8.526  8.585  8.644 

8.703  8.762   8.821    8.880  8.940 

60 

3-o 

9.000  9.060  9.120  9.181    9.242 

9-3°3   9-364  9425   9486  9.548 

62 

3-i 

9.610  9.672   9.734  9.797   9.860 

9.923   9.986    10.05    JO-11    10.18 

6 

3-2 

10.24    10.30    10.37    10.43    IO-5° 

10.56    10.63    Ia69    10.76    10.82 

7 

3-3 

10.89    10.96    11.02    11.09    11.16 

11.22    11.29    TI-36   1142    11.49 

7 

34 

11.56    11.63    :i-7°    Il'7&    JI-83 

II.9O     11-97     I2.O4     I2.II     I2.l8 

7 

3-5 

12.25    12.32    12.39    12.46    12.53 

1  2.60    12.67    I2-74    12.82    12.89 

7 

3-6 

12.96    13.03    13.10    13.18    13.25 

13.32    13.40    13.47    13.54    13.62 

7 

37 

J3-69    J3-76    13.84    13.91    13.99 

14.06    14.14    14.21    14.29    14.36 

8 

3-8 

14.44    M-S2    14-59    J4-67    14-75 

14.82    14.90    14.98    15.05    15.13 

8 

3-9 

15.21    15.29    15.37    15.44    15.52 

15.60    15.68    15.76    15.84    15.92 

8 

4.0 

16.00    16.08    16.16   16.24    16.32 

16.40    16.48    16.56    16.65    J6-73 

8 

4.1 

16.81    16.89    16.97    17.06    17.14 

17.22    17.31    17.39    J747    17-56 

8 

4.2 

17.64    17.72    17.81    17.89    17.98 

18.06    18.15    18.23    18.32    18.40 

9 

4-3 

18.49    18.58    18.66    18.75    l8-84 

18.92    19.01    19.10    19.18    19.27 

9 

44 

19.36    19.45    19.54    19.62    19.71 

19.80    19.89    19.98    20.07    20.16 

9 

4-5 

20.25    20.34    20.43    20.52    20.61 

20.70    20.79   20.88    20.98    21.07 

9 

4.6 

21.  16   21.25    21.34   21.44   21.53 

21.62    21.72    21.81    21.90   22.00 

9 

4-7 

22.09   22.18    22.28    22.37    22.47 

22.56   22.66   22.75    22.85    22.94 

10 

4.8 

23.04   23.14   23.23   23.33    23.43 

23-52    23.62    23.72.  23.81    23.91 

10 

4-9 

24.01    24.11    24.21    24.30   24.40 

24.50   24.60   24.70   24.80   24.90 

10 

5-o 
5-i 

25.00   25.10   25.20   25.30   25.40 
26.01    26.11    26.21    26.32    26.42 

25.50   25.60    25.70   25.81    25.91 
26.52    26.63    26-73    26.83    26.94 

IO 
10 

5-2 

27.04   27.14   27.25   27.35   2746 

27.56   27.67    27.77    27.88    27.98 

II 

5-3 

28.09   28.20   28.30   28.41    28.52 

28.62    28.73    28-84    28.94    29.05 

II 

54 

29.16   29.27    29.38   29.48    29.59 

29.70    29.81    29.92    30.03   30.14 

II 

n. 

01234 

56789 

Diff. 

Explanation  in  Art.  188 


TABLES 


507 


TABLE  15.    SQUARES  OF  NUMBERS 


n. 

01234 

56789 

DifT 

5-5 

3o-25   30.36   30.47    30.58   30.69 

30.80   30.91    31.02    31.14   31.25 

ii 

5-6 

31.36   31.47    31.58   31.70   31.81 

31.92    32.04    32.15    32.26   32.38 

ii 

•Si 

32.49   32.60   32.72    32.83    32.95 

33.06   33.18    33.29   33.41    33.52 

12 

5-8 

33-64   33-76   33.87    33.99   34.11 

34-22    34.34   34.46   34.57    34.69 

12 

5-9 

34.81    34.93   35.05   35.16  35.28 

35.40  35.52  35.64  35.76  35.88 

12 

6.0 

36.00   36.12   36.24   36.36   36.48 

36.60  36.72  36.84  36.97  37.09 

12 

6.1 

37-21    37.33   37.45   37.58    37.70 

37-82  37.95  38.07  38.19  38.32 

12 

6.2 

38.44   38.56   38.69   38.81    38.94 

39.06  39.19  39.31  39.44  39.56 

13 

6-3 

39.69   39.82    39.94   40.07    40.20 

40.32  40.45  40.58  40.70  40.83 

13 

6.4 

40.96   41.09   41.22    41.34   41.47 

41.60  41.73  41.86  41.99  42.12 

*3 

6.5 

42.25   42.38    42.51    42.64   42.77 

42.90  43.03  43.16  43.30  43.43 

T3 

6.6 

43-56   43-69   43-82    43-96   44-09 

44.22   44.36  44.49  44.62  44.76 

J3 

6.7 
6.8 

44.89   45.02    45.16   45.29   45.43 
46.24   46.38    46.51    46.65   46.79 

45-56   45-70  45-83   45-97    46.10 
46.92    47.06   47.20   47.33   47.47 

14 
14 

6.9 

47.61    47.75   47-89    48-02   48.16 

48.30   48.44   48.58   48.72   48.86 

H 

70 

49.00   49.14   49.28    49.42    49.56 

49.70   49.84   49.98    50.13    5027 

14 

7-i 

50.41    50.55    50.69    50.84    50.98 

51.12    51.27    51.41    51.55   51.70 

14 

7-2 

51.84    51.98    52.13    52.27    52.42 

52.56    52.71    52.85    53.00    53.14 

15 

7-3 

53.29    53.44    53.58    53.73    53.88 

54.02    54.17    54.32    54.46    54.61 

15 

74 

54.76    54.91    55.06   55.20    55.35 

55-5°   55-65   55-8o   55.95   56.10 

15 

7-5 

56.25    56.40    56.55    56.70    56.85 

57.00   57.15   57.30   57.46   57.61 

15 

7.6 

57-76   57.91    58.06    58.22    58.37 

58.52    58.68   58.83    58.98    59.14 

15 

7-7 

59.29    59.44    59.60    59.75    59.91 

60.06  60.22   60.37   60.53   60.68 

16 

60.84   61.00   61.15    61.31    61.47 

61.62   61.78   61.94  62.09  62.25 

16 

7-9 

62.41    62.57    62.73    62.88   63.04 

63.20  63.36  63.52   63.68  63.84 

16 

8.0 
8.1 

64.00   64.16   64.32    64.48   64.64 
65.61    65.77    65.93    66.10    66.26 

64.80   64.96  65.12   65.29   65.45 
66.42   66.59  66.75   66.91    67.08 

16 
16 

8.2 

67.24   67.40   67.57    67.73    67.90 

68.06  68.23   68.39  68.56  68.72 

17 

8-3 

68.89   69.06   69.22    69.39   69.56 

69.72   69.89  70.06  70.22   70.39 

17 

8.4 

70.56   70.73    70.90   71.06   71.23 

71.40   71.57   71.74   71.91    72.08 

17 

8.5 

72.25    72.42    72.59   72.76   72.93 

73.10  73.27   73.44  73.62   73.79 

17 

8.6 

73.96   74.13    74.30   74.48    74.65 

74.82   75.00   75.17    75.34   75.52 

17 

8.7 

75-69   75-86   76.04   76.21    76.39 

76.56  76.74   76.91    77.09  77.26 

18 

8.8 

77.44   77.62    77.79   77.97    78.15 

78.32   78.50   78.68   78.85   79.03 

18 

8.9 

79.21    79.39   79.57    79.74   79.92 

80.  10   80.28   80.46   80.64   80.82 

18 

9.0 

81.00   81.18   81.36   81.54   81.72 

81.90   82.08   82.26  82.45   82.63 

18 

9.1 

82.81    82.99   83.17    83.36   83.54 

83.72   83.91    84.09  84.27   84.46 

18 

9.2 

84.64   84.82    85.01    85.19   85.38 

85-56   85.75   85.93   86.12   86.30 

19 

9-3 

86.49   86.68    86.86   87.05   87.24 

87.42    87.61    87.80   87.98   88.17 

*9 

94 

88.36   88.55   88.74   88.92    89.11 

89.30   89.49   89.68   89.87   90.06 

T9 

9-5 

90.25   90.44   90.63    90.82    91.01 

91.20   91.39   91.58   91.78   91.97 

19 

9.6 

92.16   92.35   92.54   92.74   92.93 

93-12    93-32    93.51    93.70   93.90 

19 

9-7 

94.09   94.28   94.48   94.67    94.87 

95.06   95.26   95.45   95.65   95.84 

20 

9.8 

96.04   96.24    96.43   96.63   96.83 

97.02    97.22   97.42   97.67    97.81 

20 

9.9 

98.01    98.21    98.41    98.60   98.80 

99.00   99.20   99.40   99.60   99.80 

20 

«. 

01234 

56789 

Diff. 

508 


TABLES 


TABLE  16.    AREAS  OF  CIRCLES 


d 

01234 

56789 

Diff. 

.0 

.7854  .8012  .8171  .8332  .8495 

.8659  .8825  .8992  .9161  .9331 

.1 

•  95°3  -9^77  .9852   1.003  i.  021 

1.039    1.057    1-075    1.094    1.  112 

.2 

1.131   1.150  1.169  r-lS8  i'2o3 

1.227    1.247    1.267    L287    1.307 

19 

•  3 

1.327  1.348  1.368  1.389  1.410 

1.431   1.453  1-474  1-496  I.5I7 

21 

•4 

1.539  I-56j  I-584  i.  606  1.629 

1.651   1.674  1-697  1.720  1.744 

22 

•5 

1.767  1.791   1.815  1.839  1.863 

1.887  1-911   1-936  1.961  1.986 

24 

.6 

2.OII    2.036    2.O6I    2.087   2.  112 

2.138  2.164  2.190  2.217  2.243 

26 

•  7 

2.270    2.297    2.324    2.351    2.378 

2.405  2.433  2.461  2.488  2.516 

27 

.8 

2.545    2.573    2.6C2    2.630    2.659 

2.688  2.717  2.746  2.776  2.806 

29 

•9 

2.835    2.865    2.895    2.926    2.956 

2.986  3.017  3.048  3.079  3.110 

30 

2.O 

3.142    3.173    3.205    3.237    3.269 

3.301  3-333  3o65  3o98  3-431 

32 

2.1 

3.464    3.497    3.530   3.563    3.597 

3.631   3.664  3.698  3.733  3.767 

34 

2.2 

3.801    3.836   3.871    3.906    3.941 

3.976  4.012  4.047  4.083  4.119 

35 

23 

4.155    4.191    4.227    4.264   4.301 

4-337  4-374  4-412  4.449  4.486 

36 

2.4 

4.524   4.562    4.600  4.638    4.676 

4.714  4-753  4.792  4.83i  4.870 

38 

2-5 

4.909   4.948    4.988    5.O27    5.067 

5.107  5.147  5.187  5.228  5.269 

40 

2.6 

5-309  5-350  5-391  5-433  5-474 

5.5I5  5-557  5-599  5-641  5.683 

4i 

2.7 

5.726  5.768  5.811  5.853  5.896 

5.940  5.983  6.026  6.070  6.114 

43 

2.8 

6.158    6.202    6.246   6.290   6.335 

6.379  6.424  6.469  6.514  6.560 

44 

2.9 

6.605    6.651    6.697    6.743    6.789 

6.835  6.881  6.928  6.975  7.022 

46 

3-0 

7.069    7.II6    7.163    7.2II    7.258 

7.306  7.354  7.402  7-451  7-499 

48 

3-1 

7.548    7.596    7-645    7-694    7-744 

7-793  7-843  7-892  7-942  7.992 

49 

3-2 

8.042    8.OQ3    8.143    8.194    8.245 

8.296  8.347  8.398  8.450  8.501 

51 

3.3 

8.553    8.605    8.657    8.709    8.762 

8.814  8.867  8.920  8.973  9.026 

52 

3-4 

9.079   9,133   9.186   9.240   9.294 

9.348  9.402  9.457  9.511  9.566 

54 

3-5 

9.621    9.676   9.731    9.787  9-842 

9.898    9.954    IO.OI    IO.O7    IO.I2 

56 

3-6 

IO.I8    10.24    10.29    10.35    10.41 

10.46    IO.52    10.58    10.64    10.69 

6 

3-7 

10.75  10.81  10.87  IO-93  10.99 

11.04    II.  10    II.  16    11.22    11.28 

6 

3-8 

11.34  11.40  11.46  11.52  11.58 

11.64  11.70  11.76  11.82  u.88 

6 

3-9 

11.95  12.01  12.07  12.13  12.19 

12.25  12.32  12.38  12.44  12.50 

6 

4.0 

12.57  12.63  12.69  12.76  12.82 

12.88  12.95  13.01  13.07  13.14 

7 

4.1 

13.20  13.27  13  33  13.40  13.46" 

13-53  13-59  13.66  13-72  13-79 

7 

4-2 

13.85  13.92  13.99  14-05  14-12 

14.19  14.25  14.32  14.39  14-45 

7 

4-3 

14.52  14.59  14-66  14.73  14.79 

14.86  14.93  15.00  15.07  15-14 

7 

4-4 

15.21  15.27  15.34  I5.4I  15-48 

15.55  15-62  15.69  15.76  15.83 

7 

4-5 

15.90  15.98  16.05  16.12  16.19 

16.26  16.33  16.40  16.47  16.55 

7 

4.6 

16.62  16.69  16.76  16.84  16.91 

16.98  17.06  17.13  17.20  17.28 

7 

4-7 

17.35  17.42  17.50  17.57  17.65 

17.72  17.80  17.87  17.95   18.02 

8 

4.8 

18.10  18.17  18.25  18.32  18.40 

18.47  18.55  18.63  18.70  18.78 

8 

4-9 

18.86  18.93  19-01   19.09  19.17 

19.24  19.32  19.40  19.48  19.56 

8 

5.o 

19.63  19.71  19.79  19.87  19.95 

2O.O3    2O.II    2O.I9    2O.27    20.3^ 

8 

5-i 

20.43  20.51  20.59  20.67  20.75 

20.83  20.91  20.99  21.07  21.  16 

8 

5.2 

21.24  21.32  21.40  21.48  21.57 

21.65    21.73    2I.8I    21.90   21.98 

8 

5-3 

22.06  22.15  22.23  22-31  22.40 

22.48    22.56   22.65    22.73    22.82 

8 

5-4 

22.90  22.99  23.07  23.16  23.24 

23-33    23.41    23.50   23.59    23.67 

9 

</ 

01234 

56789 

Diff. 

Explanation  in  Art.  188 


TABLES 


509 


TABLE  16.    AREAS  OF  CIRCLES 


d 

01234 

56789 

Diff. 

5.5 

23.76  23.84  23.93  24.02  24.11 

24.19  24.28  24.37  24.45  24-54 

9 

5-6 

24.63  24.72  24.81  24.89  24.98 

25.07  25.16  25.25  25.34  25.43 

9 

5-7 

25.52  25.61  25.70  25.79  25.88 

25.97  26.06  26.15  26.24  26.33 

9 

5-8 

26.42  26.51  26.60  26.69  26.79 

26.88  26.97  27.06  27.15  27.25 

9 

5-9 

27.34  27.43  27-53  27.62  27.71 

27.81  27.90  27.99  28.09  28.18 

9 

6.0 

28.27  28.37  28.46  28.56  28.65 

28.75  28.84  28.94  29.03  29.13 

9 

6.1 

29.22  29.32  29.42  29.51  29.61 

29.71  29.80  29.90  30.00  30.09 

JO 

6.2 

30.19  30.29  30.39  30.48  30.58 

30.68  30.78  30.88  30.97  3r-07 

10 

6.3 

31.17  31.27  31.37  31.47  31.57 

31.67  31.77  31.87  31.97  32.07 

10 

6.4 

32.17  32.27  32.37  32.47  32.57 

32.67  32.78  32.88  32.98  33-oS 

10 

6-5 

33.18  33.29  33.39  33.49  33  59 

33.70  33.80  33.90  3400  34-11 

10 

6.6 

34.21  34.32  34.42  34.52  34.63 

34-73  34-84  34-94  35-Q5  35-15 

10 

6-7 

35.26  35.36  35.47  35.57  35.68 

35.78  35-89  36.00  36.10  36.21 

IO 

6.8 

36.32  36.42  36.53  36.64  36.75 

36.85  36.96  37.07  37.18  37-23 

II 

6-9 

37-39  37.50  37-6i  37.72  37.83 

37.94  38-05  38.16  38.26  38.37 

II 

7-o 

38.48  38.59  38.70  38.82  38.93 

39.04  39.15  39.26  39.37  39.48 

II 

7-1 

39-59  39-70  39-82  39.93  40.04 

40.15  40.26  40.38  40.49  40.60 

II 

7-2 

40.72  40.83  40.94  41.06  41.17 

41.28  41.40  41.51  41.62  41.74 

11 

7-3 

41.85  41.97  42.08  42.20  42.31 

42.43  42.54  42.66  42.78  42.89 

II 

7-4 

43.01  43.12  43.24  43.36  43.47 

43-59  43-71  43.83  43-94  44-o6 

12 

7-5 

44.18  44.30  44.41  44.53  44.65 

44.77  44.89  45.01  45.13  45.25 

12 

7.6 

45.36  45.48  45.60  45.72  45.84 

43.96  46.08  46.20  46.32  46.45 

12 

7-7 

46.57  46.69  46.81  46.93  47.05 

47-17  47-29  47.42  47.54  47-66 

12 

r  8 

47.78  47.91  48.03  48.15  48.27 

48.40  48  52  48.65  48.77  48.89 

12 

7-9 

49.02  49.14  49.27  49.39  49.51 

49.64  49.76  49.89  50.01  50.14 

12 

8.0 

50.27  50.39  50.52  50.64  50  77 

50.90  51.02  51.15  51.28  51.40 

13 

8.1 

51.53  51.66  51.78  51.91  52.04 

52.17  52.30  52.42  52.55  52.68 

13 

8.2 

52.81  52.94  53.07  53.20  53.33 

53.46  53-59  53-72  53.85  53.98 

13 

8-3 

54.11  54.24  54.37  54.50  54-63 

54.76  54.89  55.02  55.15  55.29 

13 

8.4 

55.42  55-55  55-68  55.81  55.95 

56.08  56.21  56.35  56.48  56.61 

13 

8-5 

56.75  56.88  57.01  57.15  57.28 

57-41  57.55  57.68  57.82  57.95 

13 

8.6 

58.09  58.22  58.36  58.49  58.63 

58.77  58.90  59.04  59.17  59.31 

14 

8.7 

59-45  59-53  59-72  59-s6  59-99 

60.13  60.27  60.41  60.55  60.68 

14 

8.3 

60.82  60.96  6r.io  61.24  61.38 

61.51  61.65  61.79  61.93  62.07 

14 

8.9 

62.21  62.35  62.49  62.63  62.77 

62.91  63.05  63.19  63.33  63.48 

14 

9.0 

63.62  63.76  63.90  64.04  64.18 

64.33  64.47  64.61  64.75  64.90 

14 

9.1 

65.04  65.18  65.33  65.47  65-61 

65.76  65.90  66.04  66.19  66.33 

14 

9.2 

66.48  6662  66.77  66.91  67.06 

67.20  67.35  67.49  67.64  67.78 

15 

9-3 

67.93  63.o3  63.22  63.37  68.51 

63.66  68.81  68.96  69.10  69.25 

15 

9.4 

69.40  69.55  69.69  69.84  69.99 

70.14  70.29  70.44  70.58  70.73 

15 

9-5 

70.88  71.03  71.18  71.33  7r.48 

71.63  71.78  71.93  72.08  72.23 

15 

9.6 

72.38  72.53  72.68  72.84  72.99 

73  14  73.29  73.44  73.59  73-75 

15 

9-7 

73.90  74.05  74.20  74.36  74-51 

74.66  74.82  74.97  75.12  75.28 

15 

9.8 

75-43  75.58  75.74  75.89  76.05 

76.20  76.36  76.51  76.67  76.82 

16  . 

9.9 

76.98  77.13  77-29  77-44  77-6o 

77.76  77.91  78.07  78.23  78.38 

16 

d 

01234 

56789 

Diff. 

' 

j 

510 


TABLES 


TABLE  17.    TRIGONOMETRIC  FUNCTIONS 


Angle 

Arc 

Sin 

Tap 

Sec 

Cosec 

Cot 

Cos 

Coarc 

O 

0. 

0. 

O. 

I. 

00 

CO 

I. 

1.5708 

90 

I 

0.0175 

0.0175 

0.0175 

I.O002 

57-299 

57.290 

0.9998 

•5533 

89 

2 

.0349 

•0349 

.0349 

I.  OOO6 

28.654 

28.636 

•9994 

•5359 

88 

3 

.0524 

.0523 

.0524 

I.OOI4 

19.107 

19.081 

.9986 

.5184 

87 

4 

.0698 

.0698 

.0699 

1.0024 

H.336 

14.301 

.9976 

.5010 

86 

5 

.0873 

.0872 

.0875 

1.0038 

11.474 

11.430 

.9962 

.4835 

85 

6 

o.  1047 

0.1045 

0.1051 

1-0055 

9.5668 

9.5144 

0-9945 

1.4661 

84 

7 

.1222 

.1219 

.1228 

1.0075 

8.2055 

8.1443 

.9925 

.4486 

83 

8 

.1396 

.1392 

.1405 

I.OOgS 

7.1853 

7-II54 

.9903 

•4312 

82 

9 

.1571 

.1564 

.1584 

I.OI25 

6.3925 

6.3138 

.9877 

•4137 

81 

10 

•1745 

.1736 

.1763 

I.OI54 

5.7588 

5.6713 

.9848 

.3963 

80 

ii 

0.1920 

0.1908 

0.1944 

I.OI87 

5.2408 

5.1446 

0.9816 

1.3788 

79 

12 

.2094 

.2079 

.2126 

I.O223 

4.8097 

4.7046 

.9781 

.3614 

78 

13 

.2269 

.2250 

.2309 

1.0263 

4-4454 

4.3315 

•9744 

•3439 

.  77 

14 

•2443 

.2419 

•2493 

1.0306 

4-I336 

4.0108 

.9703 

•3265 

76 

15 

.2618 

.2588 

.2679 

1-0353 

3.8637 

3.7321 

.9659 

.3090 

75 

16 

0.2793 

0.2756 

0.2867 

1.0403 

3.6280 

34874 

0.9613 

1.2915 

74 

i? 

.2967 

.2924 

.3057 

1-0457 

3.4203 

3.2709 

.9563 

.2741 

73 

18 

.3142 

.3090 

.3249 

I.05I5 

3.2361 

3-0777 

•9511 

.2566 

72 

19 

.3316 

.3256 

•3443 

1.0576 

3.0716 

2.9042 

.9455 

.2392 

7i 

20 

•3491 

.3420 

.3640 

1.0642 

2.9238 

2-7475 

•9397 

.2217 

70 

21 

0.3665 

0.3584 

0-3839 

I.07II 

2.7904 

2.6051 

0.9336 

1.2043 

69 

22 

.3840 

•3746 

.4040 

1.0785 

2.6695 

2.4751 

.9272 

.1868 

68 

23 

.4014 

.3907 

•4245 

1.0864 

2-5593 

2-3559 

.9205 

.1694 

67 

24 

.4189 

.4067 

•4452 

1.0946 

2.4586 

2.2460 

.9135 

•  1519 

66 

25 

•4363 

.4226 

.4663 

I.I034 

2.3662 

2.1445 

.9063 

•I345 

65 

26 

0.4538 

0.4384 

0.4877 

I.II26 

2.2812 

2.Q503 

0.8988 

1.1170 

64 

27 

•4712 

•4540 

•5095 

I.I223 

2.2027 

^9626 

.8910 

.0996 

63 

28 

.4887 

•4«95 

.5317 

I.I326 

2.1301 

.8807 

.8829 

.0821 

62 

29 

.5061 

.4848 

•5543 

I-I434 

2.0627 

8040 

.8746 

.0647 

61 

30 

•5236 

.5000 

•5774 

I-I547 

2.0000 

.7321 

.8660 

.0472 

60 

31 

0.5411 

0.5150 

0.6009 

1.1666 

.9416 

.6643 

0.8572 

1.0297 

59 

32 

.5585 

.5299 

.6249 

1.1792 

.8871 

.6003 

.8480 

1.0123 

58 

33 

.5760 

.5446 

.6494 

1.1924 

•8361 

•5399 

.8387 

0.9948 

57 

34 

•5934 

•5592 

.6745 

1.2062 

.7883 

.4826 

.8290 

•9774 

56 

35 

.6109 

.5736 

.7002 

1.2208 

•7434 

.4281 

.8192 

•9599 

55 

36 

0.6283 

0.5878 

0.7265 

1.2361 

.7013 

.3764 

0.8090 

0.9425 

54 

37 

.6458 

.6018 

.7536 

1.2521 

.6616 

.3270 

.7986 

.9250 

53 

33 

.6632 

.6157 

.7813 

1.2690 

.6243 

.2799 

.7880 

.9076 

52 

39 

.6807 

.6293 

.8098 

1.2868 

.5890 

•2349 

.7771 

.8901 

5i 

40 

.6981 

.6428 

.8391 

1.3054 

•5557 

.1918 

.7660 

.8727 

50 

4i 

0.7156 

0.6561 

0.^693 

1.3250 

.5243 

.1504 

0-7547 

0.8552 

49 

42 

•7330 

.6691 

.9004 

L3456 

•4945 

.1106 

•7431 

.8378 

48 

43 

.7505 

.6820 

.9325 

1-3673 

.4663 

.0724 

.7314 

.8203 

47 

44 

.7679 

.6947 

.9657 

1.3902 

.4396 

•0355 

.7193 

.8029 

46 

45 

.7854 

.7071 

i. 

1.4142 

.4142 

• 

.7071 

.7854 

45 

Coarc 

Cos 

Cot 

Cosec 

Sec 

Tan 

Sin 

Arc 

Angle 

Explanation  in  Art.  188 


TABLES 


511 


TABLE  18.    LOGARITHMS  OF  TRIGONOMETRIC  FUNCTIONS 


Angle 

Log  Arc 

Log  Sin 

Log  Tan 

Log  Sec 

Log 

Cosec 

Log  Cot 

Log  Cos 

Log 
Coarc 

0 

—  IX) 

—  oo 

—  oo 

O. 

oo 

CO 

O. 

0.1961 

90 

I 

2.2419 

2.2419 

2.2419 

0.0001 

1.7581 

1.7581 

T.9999 

.1913 

89 

2 

.5429 

.5428 

•5431 

.0003 

•4572 

•4569 

•9997 

.1864 

88 

3 

.7190 

.7188 

.7194 

.0006 

.2812 

.2806 

.9994 

.1814 

87 

4 

.8439 

•8436 

.8446 

.0011 

.1564 

-1554 

.9989 

.1764 

86 

5 

.9408 

•9403 

.9420 

.0017 

•0597 

.0580 

.9983 

.1713 

85 

6 

I.O2OO 

1.0192 

I.02I6 

0.0024 

0.9808 

0.9784 

1.9976 

0.1662 

84 

7 

.0870 

.0859 

.0891 

.0032 

.9141 

.9109 

.9968 

.1610 

83 

8 

.1450 

.1436 

.1478 

.0042 

.8564 

.8522 

.9958 

.1557 

82 

9 

.1961 

•!943 

.1997 

.0054 

.8057 

.8003 

.9946 

.1504 

81 

10 

12419 

•2397 

.2463 

.0066 

.7603 

•7537 

•9934 

.1450 

80 

ii 

Y.2833 

1.2806 

I  2887 

0.0081 

0.7194 

0.7113 

1.9919 

0.1395 

79 

I  2 

.3211 

.3179 

.3275 

.0096 

.6821 

.6725 

.9904 

.1340 

78 

13 

.3558 

•3521 

.3634 

.0113 

.6479 

.6366 

.9887 

.1284 

77 

14 

.3880 

.3837 

.3968 

.0131 

.6163 

.6032 

.9869 

.1227 

76 

15 

.4180 

•4130 

.4281 

.0151 

.5870 

.5719 

.9849 

.1169 

75 

16 

1.4460 

1.4403 

1-4575 

0.0172 

0-5597 

0.5425 

1.9828 

0.  1  1  1  1 

74 

i? 

.4723 

.4659 

.4853 

.0194 

.5341 

•5H7 

.9806 

.1052 

73 

18 

.4971 

.4900 

.5118 

.0218 

.5100 

.4882 

.9782 

.0992 

72 

19 

.5206 

.5126 

•5370 

.0243 

.4874 

.4630 

•9757 

.0931 

7i 

20 

.5429 

•5341 

.5611 

.0270 

•4659 

.4389 

•9730 

.0870 

70 

21 

1.5641 

1-5543 

1-5842 

0.0298 

0-4457 

0.4158 

1.9702 

0.0807 

69 

22 

.5843 

.5736 

.6064 

.0328 

.4264 

•3936 

.9672 

.0744 

68 

23 

.6036 

.5919 

.6279 

.0360 

.4081 

.3721 

.9640 

.0680 

67 

24 

.6221 

.6093 

.6486 

.0393 

.3907 

.3514 

.9607 

.0614 

66 

25 

.6398 

.6259 

.6687 

.0427 

•3741 

.3313 

•9573 

.0548 

65 

26 

1.6569 

1.6418 

1.6882 

0.0463 

0.3582 

0.3118 

1-9537 

0.0481 

64 

27 

.6732 

.6570 

.7072 

.0501 

.3430 

.2928 

•9499 

.0412 

63 

28 

.6890 

.6716 

.7257 

.0541 

.3284 

•2743 

•9459 

•0343 

62 

29 

•7°43 

.6856 

.7438 

.0582 

•3H4 

.2562 

.9418 

.0272 

61 

30 

.7190 

.6990 

.7614 

.0625 

.3010 

.2386 

•9375 

.O2OO 

60 

31 

1-7332 

1.7118 

1.7788 

0.0669 

0.2882 

O.22I2 

I.9331 

O.OI27 

59 

32 

.7470 

.7242 

•7958 

.0716 

•2758 

.2042 

.9284 

0.0053 

58 

33 

.7604 

.7361 

.8125 

.0764 

.2639 

.1875 

.9236 

1.9978 

57 

34 

•  7734 

.7476 

.8290 

.0814 

.2524 

.I7IO 

.9186 

.9901 

56 

35 

.7859 

.7586 

.8452 

.0866 

.2414 

.1548 

.9134 

.9822 

55 

36 

1.7982 

1.7692 

1.8613 

0.0920 

0.2308 

0.1387 

1.9080 

1-9743 

54 

37 

.8101 

•7795 

.8771 

.0977 

.2205 

.1229 

.9023 

.9662 

53 

38 

.8217 

.7893 

.8928 

.1035 

.2107 

.1072 

.8965 

•9579 

52 

39 

.8329 

.7989 

.9084 

.1095 

.2011 

.0916 

.8905 

•9494 

5i 

40 

•8439 

.8081 

•9238 

.1157 

.1919 

.0762 

•8843 

.9408 

50 

4i 

1.8547 

1.8169 

1.9392 

0.1222 

0.1831 

0.0608 

1.8778 

I.932I 

49 

42 

.8651 

.8255 

-9544 

.1289 

.1745 

.0456 

.8711 

.9231 

48 

43 

•8753 

•8338 

.9697 

•1359 

.1662 

.0303 

.8641 

.9140 

47 

44 

.8853 

.8418 

.9848 

.1431 

.1582 

.0152 

.8569 

.0046 

46 

45 

•  8951 

.8495 

o. 

.1505 

.1505 

O. 

•  8495 

-8951 

45 

Log 
Coarc 

Log  Cos 

Log  Cot 

Log 
Cosec 

Log  Sec 

Log  Tan 

Log  Sin 

Log  Arc 

Angle 

Explanation  in  Art.  188 


512 


TABLES 


TABLE  19.    LOGARITHMS  OF  NUMBERS 


n 

01234 

56789 

Diff. 

10 

oooo  0043  008^  0128  0170 

0212   0253   0294   0334   0374 

42 

n 

0414  0453  0492  0531  0569 

0607   0645   0682   0719   0755 

38 

12 

0792  0828  0864  0899  0934 

0969  1004  1038  1072  1106 

35 

13 

1139  1173  1206  1239  1271 

T3°3  T335  T367  '399  '43° 

32 

1461  1492  1523  1553  1584 

1614  1644  J673  1703  1732 

30 

15 

1761  1790  1818  1847  1875 

1903  1931  1959  1987  2014 

28 

16 

2041   2068   2095   2122   2148 

2175  2201  2227  2253  2279 

27 

17 

2304   2330   2355   2380   2405 

2430  2455  2480  2504  2529 

25 

18 

2553   2577   26OI   2625   2648 

2672  2695  2718  2742  2765 

24 

19 

2788   2810   2833   2856   2878 

2900  2923  2945  2967  2989 

22 

20 

3010   3032   3054   3075   3096 

3IlS  3139  3l6o  3181  3201 

21 

21 

3222   3243   3263   3284   3304 

3324  3345  3365  3385  3404 

2O 

22 

3424  3444  3464  3483  3502 

3522  354i  3560  3579  3598 

19 

23 

3617  3636  3655  3674  3692 

3711  3729  3747  3766  3784 

18 

24 

3802  3820  3838  3856  3874 

3892  3909  3927  3945  3962 

18 

25 

3979  3997  4014  4031  4048 

4065  4082  4099  4116  4133 

17 

26 

4150  4166  4183  4200  4216 

4232  4249  4265  4281  4298 

17 

27 

4314  4330  4346  4362  4378 

4393  4409  4425  4440  4456 

16 

28 
29 

4472  4487  4502  4518  4533 
4624  4639  4654  4669  4683 

4548  4564  4579  4594  4609 
4698  4713  4728  4742  4757 

15 

30 

4771  4786  4800  4814  4829 

4843  4857  4871  4886  4900 

14 

31 

4914  4928  4942  4955  4969 

4983  4997  5011  5°24  5°38 

14 

32 

505i  5°^  5079  5092  5105 

5IT9  5T32  5T45  5r59  5T72 

13 

33 

5185  5198  5211  5224  5237 

5250  5263  5276  5289  5302 

13 

34 

53i5  5328  5340  5353  5366 

5378  5391  5403  5416  5428 

13 

II 

544i  5453  5465  5478  5490 
5563  5575  5587  5599  5611 

5502  5514  5527  5539  555i 
5623  5635  5647  5658  567o 

12 
12 

37 

5682  5694  5705  5717  5729 

5740  5752  5763  5775  5786 

12 

38 

5798  5809  5821  5832  5843 

5855  5866  5877  5888  5899 

II 

39 

5911  5922  5933  5944  5955 

5966  5977  5988  5999  6010 

II 

40 

6021  6031  6042  6053  6064 

6075  6085  6096  6107  6117 

II 

41 

6128  6138  6149  6160  6170 

6180  6191  6201  6212  6222 

II 

42 

6232  6243  6253  6263  6274 

6284  6294  6304  6314  6325 

IO 

43 
44 

6335  §345  6355  6365  6375 
6435  6444  6454  6464  6474 

6385  6395  6405  6415  6425 
6484  6493  6503  6513  6522 

IO 
10 

46 

6532  6542  6551  6561  6571 
6628  6637  6646  6656  6665 

6580  6590  6599  6609  6618 
6675  6684  6693  6702  6712 

IO 

9 

47 

6721  6730  6739  6749  6758 

6767  6776  6785  6794  6803 

9 

48 

6812  6821  6830  6839  6848 

6857  6866  6875  6884  6893 

9 

49 

6902  6911  6920  6928  6937 

6946  6955  6964  6972  6981 

9 

50 

6990  6998  7007  7016  7024 

7033  7042  7050  7059  7067 

9 

7076  7084  7093  7101  7110 

7118  7126  7135  7143  7152 

8 

52 

7160  7168  7177  7185  7193 

7202  7210  7218  7226  7235 

8 

53 

7243  7251  7259  7267  7275 

7284  7292  7300  7308  7316 

8 

54 

7324  7332  7340  7348  7356 

7364  7372  7380  7388  7396 

8 

H 

o    r     2    3    4 

56789 

Diff. 

Explanation  in  Art.  188 


TABLES 


513 


TABLE  19.    LOGARITHMS  OF  NUMBERS 


n 

01234 

56789 

Diff. 

55 

7404  7412  7419  7427  7435 

7443  745T  7459  7466  7474 

8 

56 

7482  7490  7497  7505  7513 

7520  7528  7536  7543  7551 

I 

7559  7566  7574  7  582  7589 
7634  7642  7649  7657  7664 

7597  7604  7612  7619  7627 
7672  7679  7686  7694  7701 

S9 

7709  7716  7723  7731  7738 

7745  7752  776o  7767  7774 

60 

7782  7789  7796  7803  7810 

7818  7825  7832  7839  7846 

7 

61 

7853  7860  7868  7875  7882 

7889  7896  7903  7910  7917 

62 

7924  793  i  793s  7945  7952 

7959  7966  7973  798o  7987 

63 

7993  8000  8007  8014  8021 

8028  8035  8041  8048  8055 

64 

8062  8069  8075  8o82  8089 

8096  8102  8109  8116  8122 

65 

8129  8136  8142  8149  8156 

8162  8169  8176  8182  8189 

7 

66 

8195  8202  8209  8215  8222 

8228  8235  8241  8248  8254 

67 

8261  8267  8274  8280  8287 

8293  8299  8306  8312  8319 

68 
69 

832=,  8331  8338  8344  8351 
8388  8395  8401  8407  8414 

8357  8363  8370  8376  8382 
8420  8426  8432  8439  8445 

70 

8451  8457  8463  8470  8476 

8482  8488  8494  8500  8506 

6 

71 

85*3  8519  8525  8531  8537 

8543  8549  8555  8561  8567 

72 

8573  8579  8585  8591  8597 

8603  8609  8615  8621  8627 

73 

8633  8639  8645  8651  8657 

8663  8669  8675  8681  8686 

74 

8692  8698  8704  8710  8716 

8722  8727  8733  8739  8745 

75 

8751  8756  8762  8768  8774 

8779  8785  8791  8797  8802 

6 

76 

8808  8814  8820  8825  8831 

8837  8842  8848  8854  8859 

77 

8865  8871  8876  8882  8887 

8893  8899  8904  8910  8915 

78 

8921  8927  8932  8938  8943 

8949  8954  8960  8965  8971 

79 

8976  8982  8987  8993  8998 

9004  9009  9015  9020  9025 

80 

9031  9036  9042  9047  9053 

9058  9063  9069  9074  9079 

5 

81 

9085  9090  9096  9101  9106 

9112  9117  9122  9128  9133 

82 

9138  9T43  9M9  9T54  9*59 

9165  9170  9175  9180  9186 

83 

9191  9196  9201  9206  9212 

9217  9222  9227  9232  9238 

84 

9243  9248  9253  9258  9263 

9269  9274  9279  9284  9289 

85 

9294  9299  9304  93°9  93  T  5 

9320  9325  9330  9335  9340 

5 

86 

9345  9350  9355  936o  9365 

9370  9375  938o  9385  9390 

87 

9395  9400  9405  9410  9415 

9420  9425  9430  9435  9440 

88 

9445  9450  9455  946o  9465 

9469  9474  9479  9484  9489 

89 

9494  9499  95°4  95O9  95  '3 

95'8  9523  9528  9533  953* 

90 

9542  9547  9552  9557  9562 

9566  9571  9576  9581  9586 

5 

91 

9590  9595  96oo  9605  9609 

9614  9619  9624  9628  9633 

92 

9638  9643  9647  9652  9657 

9661  9666  9671  9675  9680 

93 

9685  9689  9694  9699  9703 

9708  9713  9717  9722  9727 

94 

9731  9736  974i  9745  9750 

9754  9759  9763  9768  9773 

95 

9777  9782  9786  9791  9795 

9800  9805  9809  9814  9818 

4 

96 

9823  9827  9832  9836  9841 

9845  9850  9854  9859  9863 

98 

9868  9872  9877  988  r  9886 
9912  9917  9921  9926  9930 

9890  9894  9899  9903  9908 
9934  9939  9943  994°  9952 

99 

9956  996i  9965  9969  9974 

9978  9983  9987  9991  9996 

n 

01234 

56789] 

Diff. 

514 


TABLES 


TABLE  20.    CONSTANTS  AND  THEIR  LOGARITHMS 


Name. 
(Radius  of  circle  or  sphere  =  i.) 

Symbol. 

Number. 

Logarithm. 

Ajee?Tof  circle 

it 

3   141  592654 

0.497  149873 

Circumference  of  circle 

in 

6.283  185  307 

0.798  179868 

Surface  of  sphere 

47t 

12.566  370614 

1.099209  864 

** 

0.523  598  776 

1.718998622 

Quadrant  of  circle 

& 

0.785398  163 

1.895089881 

Area  of  semicircle 

fr 

1.570796327 

0.196  119877 

Volume  of  sphere 

ire 

4.188  790  205 

O.622  088  609 

7T2 

9.869604401 

0.994299745 

** 

1.772453851 

0.248  574936 

Degrees  in  a  radian 

l80/7T 

57-2957795I3 

1.758  122632 

Minutes  in  a  radian 

I0800/7T 

3437.746771 

3.536273883 

Seconds  in  a  radian 

648ooo/7T 

206  264  .  806 

5.314425  133 

I/*. 

0.318  309886 

1.502  850  127 

I/** 

0.564  189  584 

1.751425064 

I/*2 

o.ioi  321  184 

1.005  700255 

Circumference/36o 

arc    ° 

0.017453293 

2.241  877368 

sin    ° 

0.017  452406 

2.241  855318 

Circumference/2  1600 

arc    ' 

0.000290888 

4.463  726  117 

sin    ' 

0.000290888 

4.463  726  III 

Circumference/  1  296000 

arc    " 

0.000004848 

6.685574867 

sin    " 

o  .  ooo  004  848 

6.685  574867 

Base  Naperian  system  -of  logs 

e 

2.718  281  828 

0.434294482 

Modulus  common  system  of  logs 

M 

0.434294482 

1.637784311 

Naperian  log  of  10 

i/M 

2.302  585093 

0.362  215  689 

hr 

0.4769363 

1.6784604 

Probable  error  constant 

hr  f/2 

0.6744897 

1.8289754 

Feet  in  one  meter 

m/ft. 

3.2808333 

0.5I5984I 

Miles  in  one  kilometer 

km/mi. 

0.621  369  9 

1-7933502 

INDEX 


515 


INDEX 


Absorption  of  brick,  64 

Acid  steel,  61 

Alloys,  65,  67 

Aluminum,  66 

American  Society  for  Testing  Materials, 

478,  482,  486 
Angles,  104,  no,  in,  502 
Angular  velocity,  421 
Annealing,  59,  64,  305 
Anthracite  coal,  67,  380 
Answers  to  problems,  493 
Apparent  and  true  stresses,  186,  274, 

359-382,  460 

Approximate  computations,  34 
Area,  reduction  of,  31 
Areas  of  circles,  21,  508 
Army,  gun  formulas,  394-402 
Artificial  stone,  54,  66 
Association,  testing,  486 
Axial  stresses,  i,  3,  190,  253,  327,  365 

impact,  327,  331,  337 
Axis,  neutral,  98,  100 

of  a  bar,  2,  188 

of  a  beam,  100,  290 

of  a  column,  190 

of  a  section,  439 
Axle  steel,  63 
Axles,  348,  353 

Bach,  C.,  414,  416 

Bar,  i,  42,  69 

Bar  iron,  58,  505 

Bars  of  uniform  strength,  71 

resilience  of,  306 

under  centrifugal  stress,  421 

under  impact,  327,  331 

weights  of,  2,  494 
Base-line  apparatus,  165 
Basic  steel,  61 
Bauschinger,  J.,  353 


Beams,  87-187,  253-267,  269-302 
bending  moments,  93,  116 
cantilevers,  116-148,  503 
cast  iron,  122,  128 
center  of  gravity  of  sections,  103 
centrifugal  stress,  425 
combined  stresses,  251-275 
concrete,  282-298 
constrained,  149-167,  503 
continuous,  87,  168-187 
curved,  433 
deck,  no,  500 
definitions,  87 
deflection,   112,   135,  145,  153,  258, 

312-319,  503 

deflection  and  stiffness,  142,  158 
deflection  and  stress,  143,  159 
designing  of,  125,  292 
elastic  curve,  87,  114,  136,  138 
elastic  resilience,  308 
experimental  laws,  99,  185 
fixed,  149-167,  503 
flexural  strength,  56,  131 
flitched,  282 

fundamental  formulas,  101,  102,  185 
Galileo's  investigations,  186,  470 
historical  notes,  184,  470 
horizontal  shear,  269 
impact  on,  329,  334 
influence  lines,  431 
internal  stresses,  97,  270 
internal  work,  303-323 
lines  of  stress,  272 
maximum  moments,  119,  150,  154 
modulus  of  rupture,  47,  131 
moments  of  inertia,  105-108 
moving  loads,  132 
plate-girder,  108,  247,  298 
pure  flexure,  374 
overhanging,  149-165 


516 


INDEX 


Beams,  reactions,  88,  150,  169 

reinforced-concrete,  285,  298 

resilience  of,  308 

rolled,  108 

safe  loads  for,  124 

simple,  116-148,  503 

stiffness,  141,  158 

sudden  loads,  324 

theoretical  laws,  97,  185 

true  stresses,  367 

uniform  strength,  143-148 

unsymmetric  loads,  427 

vertical  shear,  90,  161 

weights  of.  42,  88 
Bearing  compression,  85 
Bending  moment,  93,  98,  101,  116 

diagrams  of,  98,  116 

influence  lines,  431 

maximum,  119,  150,  503 

maximum  maximorum,  133 

tables  of,  175,  503 

triangular  load,  426 
Bessemer  steel,  60,  305 
Best  iron,  57 

Bethlehem  Steel  Co.,  64,  241,  383 
Beton,  66 

Birnie's  formulas,  394 
Boiler  steel,  63,  482 
Boilers,  83,  417 

joints  in,  81-86 

tubes  in,  78 

Bolts,  16,  239,  265,  366 
Brass,  67 
Bresse,  M.,  182 
Brick,  6,  24,  44,  48,  252,  380 

strength  of,  13,  14,  17,  49,  131,  506 

weight  of,  42,  48 
Brick  masonry,  49 
Brick  tower,  50 

Bridges,  17,  40,  59,  86,  187,  262,  284 
Bridge  iron,  58 

rollers,  404 
steel,  482 
Briquettes,  53 

Brittle  materials,  44,  380,  476 
Brittleness,  6,  43 
Bronze,  67 
Buildings,  unit-stresses,  18.  498 


Building  stone,  51 

Bulb  beams,  no,  427,  498 

Butt-joints,  82,  85 

Campbell,  H.  H.,  61 
Cantilever  beams,  87,  116-148,  503 
deflection  of,  135,  145 
elastic  curve,  136 
fundamental  formulas,  102 
internal  work,  308 
resilience,  303 
table  for,  503 
uniform  strength,  144 
with  constraint,  163 
Carbon  in  cast  iron,  55 
steel,  60,  62 
wrought  iron,  58 
Castings,  55,  65,  482 
Cast  iron,  55,  496-498 
beams,  128 
brittleness  of,  44,  380 
elastic  limit,  5,  56,  496 
factors  of  safety,  17 
flexural  strength,  56,  131 
in  compression,  13,  56 
in  shear,  14,  38 
in  tension,  10,  24,  56 
pipes,  76 
resilience  of,  306 
weight  of,  42,  55 
Cement,  52,  54 

testing  of,  475 
Center  of  gravity,  73,  103 

of  gyration,  112 
Centrifugal  stress,  4.21,  425 
Chain  link,  502 
Channels,  104,  no,  501 
Chestnut,  46,  47 
Christie's  experiments,  197 
Circles,  areas  of,  106,  508 
Circular  plates,  .109,  411 

rings,  476 

Clapeyron,  E.,  174,  182 
Classification  of  pig  iron,  55 

steel,  63 

Clavarino's  formulas,  393 
Coal,  67,  380 
Coefficient  of  elasticity,  24 


INDEX 


517 


Coefficient  of  expansion,  252,  506 

of  impact,  350 

of  inertia,  331 

of  internal  friction,  378,  380 
Cold  bend  test,  58,  63,  439,  481 

rolling,  58 
Columns,  12,  188-224,  279 

compound,  276-281 

deflection  of,  194 

design  of,  206,  219 

eccentric  loads,  214,  217 

ends  of,  191,  221 

Euler's  formula,  192 

experiments  on,  196 

Gordon's  formula,  203,  208 

Hodgkinson's  formula,  197 

investigation  of,  203,  219,  279 

Johnson's  (T.  H.)  formula,  208 

radius  of  gyration,  191 

Rankine's  formula,  200,  211 

reinforced-concrele,  279 

Ritter's  formula,  211,  221 

rupture  of,  197,  213 

safe  loads,  for,  205 

sections  of,  188,  281 

theory  of,  190,  220 
Combined  stresses,  251-275 

compression  and  flexure,  254 

flexure  and  torsion,  259 

shear  and  tension,  264 

tension  and  compression,  251 

tension  and  flexure,  259 

torsion  and  compression,  268 
Comparison  of  beams,  141,  158,  503 
Compound  beams,  282-302 

columns,  276-281 
cylinder,  390 
Compression,  2,  n,  31,  1 88,  478,  497 

and  flexure,  255,  262 

and  shear,  265 

and  tension,  251 

and  torsion,  266 

cast  iron,  14,  17,  55 

cement,  53,  438 

concrete,  54,  288 

eccentric  loads,  72,  214,  217 

mortar,  52 

on  rivets,  81,  366 


Compression^  steel,  14,  17,  60 

stone.  14,  51 

wrought  iron,  14,  17,  44,  57 
Compressive  tests,  45,  478 
Computations,  19,  34 
Concentrated  loads,  88,  119,  407 
Concentric  loads,  190,  368 
Concrete,  54,  279,  288,  498 
Concrete  beams,  282-302 

columns,  279 
Connecting  rod,  426 
Constants,  tables  of,  496-514 
Constrained  beams,  149 
Continuity,  168 
Continuous  beams,  87,  168-187 

equal  spans,  175 

fixed  ends,  179 

influence  lines,  432 

properties  of,  171 

tables  of,  175,  176 

three  moments,  1 73 

unequal  spans,  177 
Contraction  of  area,  31 
Cooper,  T.,  85,  210,  404 
Cox,  H.,  342 
Couplings  for  shafts,  239 
Crandall,  C.  L.,  407 
Crank  arm,  242,  244 

pin,  241,  243,  459 
Crehore,  J.  D.,  212 
Crucible  steel,  60 
Cubic  equation,  166,  460 
Curvature,  radius  of,  114 
Curved  beams,  433 
Crystals  in  steel,  354,  382 
Cylinders,  77,  383 

compound,  390,  399 

exterior  pressure,  77 

interior  pressure,  75,  383 

thick,  76,  383-395 

thin,  75,  77,  394 

with  hoops,  78,  383,  309 
Cylindrical  rollers,  403 

Dead  loads,  132,  349 
Deck  beams,  no,  500 
Deflection  of  beams,  112,  312-319,  503 
cantilever  beams,  135, 145 


518 


INDEX 


Deflection  of  compound  beams,  300 

constrained  beams,  151-156 

simple  beams,  138,  147,  301 

sudden  loads,  324 

under  impact,  329,  334 

under  moving  load,  350 

under  shearing,  302,  317 
Deflection  of  columns,  194,  218 
of  plate  girders,  301 
Deformation,  elastic,  3,  8,  23,  28 
ultimate,  30 
work  in,  35 
Designing,  18 

beams,  125,  292 

columns,  206 

guns,  384,  389,  399 

joints,  83 

shafts,  231,  241 
Detrusion,  38,  91 
Diagram  of  stress,  9,  28 
Diagrams,  shear  and  moment,  92,  94, 

116,  150,  173 

Dimensions  in  equations,  21 
Ductility,  n,  58,  63 
Dudley,  C.  B.,  487 
Dudley,  P.  H.,  343 
Dynamic  stress,  324-358 

Eccentric  loads,  72-74,  214,  216,  217 

Economic  beams,  1 29 

Economy  in  design,  18,  127 

Efficiency  of  a  joint,  82 

Elastic  curve,  87,  112,  151,  153,  156 

cantilever  beams,  135 

columns,  193,  218 

constrained  beams,  141-149 

continuous  beams,  173 

general  equation,  112 

simple  beams,  139 
Elastic  deflection,  112 
Elastic  deformation,  23-37,  448 
Elastic  limit,  4,  9,  17,  23,  27,  496 

cast  iron,  5,  56 

compression,  13 

shear,  15,  39 

steel,  5,  27,  63 

tension,  5,  477 

timber,  5,  47 


Elastic  limit,  wrought  iron,  5,  58 
Elastic  resilience,  303-317 
Elastic  strength,  1-32 
Elasticity,  coefficient  of,  24 
laws  of,  4,  8,  45 
modulus  of,  23,  464 
theory  of,  447-469 
Electric  analogies,  278,  490 
Ellipse  of  stress,  461 
Ellipsoid  of  stress,  454 
Elliptical  plates,  414 
Elongation,  3,  4,  10,  26 

ultimate,  10,  30,  56,  58,  63,  482 

under  impact,  325,  33 2 

under  own  weight,  70 
Endurance  tests,  358 
Energy,  303,  306,  445,  467 
Equations,  dimensions,  21 
Equilibrium,  2,  96 
Ether,  470,  473 
Euler's  formula,  192,  196 
Expansibility,  251,  480 
Experimental  laws,  8,  17,  99,  225,  360 
External  forces,  i,  39,  77 
External  work,  35,  303,  312,  324 
Eye  bars,  260,  445 

Fairbairn,  W.,  78 

Factor  of  lateral  contraction,  34,  359 

of  safety,  7,  8,  17 
Fatigue  of  materials,  352~358,  381 
Fixed  beams,  87,  149,  152,  156 
Flexural  strength,  47,  56,  131,  497 
Flexure,  87-186 

and  compression,  255,  262 

and  tension,  259,  262 

and  torsion,  266 

centrifugal,  425 

curved  beams,  434 

effect  of  bends,  436 

erroneous  views,  185 

formula,  101 

of  crank  pin,  241 

of  joints,  82,  86 

pure,  374 

under  impact,  329,  334 

under  live  load,  132,  349 

work  of,  308,  312 


INDEX 


519 


Floor  beams,  147,  300 
Flues,  boiler,  78 
Fly  wheel,  423 
Forge  pig,  55,  57 
Forgings,  64 
Foundry  pig,  55 
Friction,  internal,  375-382 

Galileo,  G.,  186,  470 
German  I  beams,  109,  504 
Glass.  67 

Gordon,  L.,  203,  208 
Goss,  W.  F.  M.,  485 
Granite,  50,  51 
Gravitation,  468]  489 
Gravity,  center  of,  73,  103 

specific,  42 

Grecian  columns,  223 
Greek  letters,  21 
Grindstone,  424 
Gun  metal,  63 
Guns,  34,  383-401 

hooped,  384,  390,  399 

solid,  388,  393 
Gyration,  radius  of,  112,  129 

Hard  steel,  62 

Hartmann,  L.,  376 

Hatt,  W.  K.,  347,  348,  487 

Helical  springs,  445 

Hemispheres,  417,  419 

Hemlock,  46 

Historical  notes,  39,  186,  196,  318,  341, 

433,  4/0 

Hodgkinson,  E.,  342,  347 
Hollow  cylinders,  75,  383-401 

shafts,  232 

spheres,  77 

Hooke's  law,  4,  40,  449 
Hoops,  centrifugal  stress,  422 

for  guns,  383,  391,  396,  399 

shrinkage  of  79,  396 
Horizontal  impact,  330,  336 
shear,  269 
stresses,  97,  100 
Horse-power,  230 
Howard,  J.,  358,  473 


Hydraulic  cement,  52 
mortar,  53 

I  beams,  103,  107,  122,  124,  128,  178, 

483,  488 
Impact,  324-351 
on  bars,  327,  331 
on  beams,  329,  334 
pressure  due  to,  344 
tests  on,  341,  346,  481 
Inertia  of  a  bar,  331 

of  a  beam,  334 
in  impact,  33!~337 
moment  of,  105,  438-442 
product  of,  437 
Inflection  point,  149 

friction,  375-382 
Influence  lines,  431 
Inspection  of  material,  485 
Internal  stresses,  2,  96,  35  ,  467 

work,  303-315 

International  Association,  486 
Investigation,  17,  122,  203 
of  beams,  121 
of  columns,  203 
of  guns,  389,  393,  401 
of  joints,  80 
of  shafts,  230,  233 
Iron,  44,  55,  58 
Isotropic  materials,  447 

Jacket  for  guns,  384 
Johnson,  J.  B.,  368,  409 
Johnson,  T.  H.,  208 
Joints,  riveted,  80-86 

Keep,  W.  J.,  342 

Keep's  impact  machine,  343 

Kirkaldy,  D.,  341 

Lame,  E.,  395,  478 

Lame's  formulas,  385 

Lap  joints,  80,  84 

Lateral  contraction,  32,  34,  359 

factor  of,  34 

Launhardt's  formula,  355 
Laws,  experimental,  8,  17,  99 
of  fatigue,  353 


520 


INDEX 


Laws  of  internal  stresses,  360 

of  resilience,  303,  324 
Lead,  67 
Least  work,  320 
Lilly's  column  formula,  223 
Limestone,  50,  51,  482 
Limiting  length  of  bar,  69 

of  beam,  167 
Live  loads,  132,  349,  470 
Loads,  3,  87,  407 

safe,  for  beams,  124 

safe,  for  columns,  205 

sudden,  324 
Locomotive,  425 
Log,  beam  cut  out,  129 
Logarithms,  511-514 
Long  columns,  192 

Machinery  steel,  63 

Malleable  cast  iron,  57 

Manhole  covers,  415 

Marburg,  E.,  212,  487 

Marston,  A.,  407 

Martens,  A.,  435 

Masonry,  brick,  49 
stone,  51 

Masonry  piers,  214,  473 

Materials,  factors  of  safety,  7-17 
fatigue  of,  352-358 
properties  of,  42-68 
resilience  of,  303-320 
specifications  for,  482 
strength  of,  1-22,  42-68 
tests  of,  472-487 
weights  of,  42,  496 

Maximum  moments,  119,  133,  432,  503 
shears,  119,  134,  432,  457 

Measures,  systems  of,  20,  21 

Medium  steel,  62,  63,  482 

Meigs,  J.  F.,  401 

Merriman,  M.,  182,  221,  356 

Metric  system,  20 

Millstone,  424 

Modulus  of  elasticity,  23,  37,  47,  496 
resilience,  307 
rupture,  47,  132,  497 

Moisture  in  timber,  47 

Moment  of  a  force,  93 


Moment  of  bending,  94 

twisting,  226 
Moment  of  inertia,  105,  in,  438 

for  beams,  105,  108,  429,  431,  503 

for  columns,  189 

for  shafts,  229 
Moments,  bending,  93,  in,  116 

cantilever  beams,  94,  117 

continuous  beams,  176 

diagrams  of,  116 

fixed  beams,  152,  156 

influence  lines,  431 

maximum,  119,  133 

overhanging  beams,  150 

resisting,  98 

simple  beams,  95,  118 

theorem  of  three,  173 
Moncrieff,  J.  M.,  224 
Mortar,  52 
Moving  loads,  132,  349 

Natural  cement,  52,  54 
Navier,  L.  M.  H.,  186 
Navy,  gun  formulas,  393,  402 
Neutral  axis,  100,  429 

surface,  99,  434 
Newton,  L,  8,  40,  382,  494 
Nickel  steel,  65 
Normal  stress,  264,  362,  450 
Norton,  W.  A.,  318 

Oak,  46,  47,  482 
Olsen,  testing  machine,  472 
One-hoss  shay,  18 
Open-hearth  steel,  60,  61,  482 
Ordnance  formulas,  383-401 
Ores  of  iron,  57,  60 
Oscillations  of  a  bar,  325,  333 

of  a  beam,  344 

Overhanging  beams,  149-155,  165,  433 
Own  weight  of  a  bar,  69 

of  beam,  124 

Parabola,  94,  118,  144,  357 
Parallel  rod,  425 
Paving  brick,  49 
Phosphor  bronze,  66 
Phosphorus  in  steel,  61,  482 


INDEX 


521 


{Pitch  of  rivets,  81,  85 
Piers,  71,  216 
Pig  iron,  55 
Piles,  279,  281  1 
Pine,  46,  47,  318 
Pipes,  76,  383,  388 

thick,  383-392 

thin,  75,  390 
Piston  rod,  19,  188 
Plasticity,  43 

Plate  girder,  108,  147,  298,  369 
Plates,  58,  63,  403-416,  474,  485 

on  cylinder  ,  83,  419 
Poisson's  ratio,  34 
Polar  moments  of  inertia,  229,  439 
Portland  cement,  52,  54 
Powder  for  guns,  383 
Power,  shafts  for,  230 
Pressure  due  to  impac  ,  345 
Principal  axes,  440 

stresses,  455 

Prisms,  loads  on,  2,  74,  214 
Problems,  answers  to,  493 
Product  of  inertia,  437 
Puddling  furnace,  58 
Pure  stresses,  373 
Purlins,  428 

Radius  of  curvature,  114 

of  gyration,  112,  129,  191 
Rafters,  254 

Railroad  rails,  43,  104,  343 
Range  of  stresses,  352-358 
Reactions  of  beams,  88,  91 
Rectangle,  105 
Rectangular  beams,  129,  288 
plates,  415 
shafts,  247 

Reduction  of  area,  31 
Reinforced  concrete,  279-298 
Reinforcing  plates,  282 

rods,  286,  297 
Rejto,  A.,  378 
Repeated  stresses,  352-358 
Resilience,  303-318,  467 

of  bars,  304,  306 

of  beams,  308 

of  shearing,  310 


Resilience  of  torsion,  311 
Resisting  moment,  98,  227 

shear,  98,  101 
Resultant  stress,  452 
Ring,  circular,  492 
Ritter,  A.,  211,  221 
Riveted  joints,  80-86 

design,  of,  83 

efficiency  of,  82 
Rivet  iron,  58 

steel,  63,  445 
Rivets,  80-85,  483,  498 
Rolled  beams,  109 
shapes,  no 
Rollers,  403,  406 
Ropes,  66,  278 
Round  shafts,  229,  231 
Rupture,  3,  8,  15,  31,  45,  475 

beams,  130 

columns,  197 

modulus  of,  131,  497 

reinforced  concrete  beams,  294 

repeated  stress,  352 

Safe  loads,  17,  124 
Safety,  factors  of  7,  17 
Saint  Venant,  B.  de,  249,  448 
Sand-lime  brick,  66 
Sandstone,  50-51,  380 
Section,  changes  in,  32 
Section  area,  i,  42 
factor,  102 
Set,  5,  28,  29 
Shafts,  225-250,  266,  268 

couplings  for,  239 

cranks  for,  241,  243] 

for  power,  230 

hollow,  232 

round,  227,  231 

square,  248 

resilience  of,  310 

stiffness  of,  231 

strength  of,  231,  237 

true  stresses,  371 
Shapes,  rolled,  107,  no,  499-504 
Shear,  14,  15,  369,  457 

and  tension,  15,  263,  362 

deflection  due  to,  316 


522 


INDEX 


Shear,  horizontal,  269 

influence  lines,  431 

on  rivets,  81 

resilience  of,  310 

resisting,  98 

vertical,  90,  98 

ultimate  strength,  14,  370 

work  of,  39,  310 
Shear  formula,  101 
Shearing  modulus,  37,  234,  463 

strength,  497 
Shears  for  cantilevers,  91,  117 

for  continuous  beams,  176 

for  simple  beams,  92,  118,  120 
Shocks,  17,  57 
Shortening,  3 

Shrinkage  of  hoops,  79,  393 
Simple  beams,  87,  89,  116-148,  503 
Slate,  51,  482 
Slenderness  ratio,  191 
Soft  steel,  62,  445 
Solid  shafts,  229,  231,  237 
Sound,  velocity  of,  489 
Specific  gravities,  42 
Specifications,  18,  477,  482 
Specimens  for  tests,  474 
Spheres,  77,  417 
Spherical  rollers,  406 
Spiral  springs,  443 
Spring  steel,  63 
Springs,  341,  442 
Square  plates,  416 

shafts,  248,  480 
Squares  of  numbers,  485,  506 
Stability,  18,  127 
Static  loads  and  stresses,  324 
Steam  boilers,  76,  78 

pipes,  75 

Steel,  5,  10,  13,  17,  24,  60-65,  132,  252, 
354,  482 

constants  of,  496-498 

factors  of  safety,  17 

properties  of,  60-65 
1  resilience,  307 

weight  of,  42 

Steel  beams,  103,  107,  124,  499,  504 
cranks,  241,  243 
guns,  384 


Steel  plates,  410,  413,  474,  485 
pipes,  76 
rollers,  404 
ropes,  278 
spheres,  407,  419 
Stiffness  of  beams,  141,  158,  503 

of  shafts,  237 

Stone,  13,  14,  17,  42,  44,  50,  131,  498 
Straight-line  formula,  208 
Strength  of  materials,  1-22,  42-68 
history  of,  39,  186 
tables,  480,  481 
Stress,  i,  42,  100,  139,  447 
apparent,  275 
centrifugal,  421,  424 
combined,  251-275 
diagrams  of,  9,  28,  29 
in  guns,  383-402 
pure,  373 
repeated,  356 
sudden,  324 
temperature,  68,  251 
true,  359-362 
working,  17 
Stringer,  284 
Strong  steel,  5,  n,  13 
Structural  steel,  5,  10,  u,  13,  14,  17, 

63,  109,  128,  482 
Sudden  deflections,  325 

loads,  324 

Supports  of  beams,  87,  89,  159 
Surface,  neutral,  98 

T  shapes,  104,  107,  in,  128,  500 
Tables,  20,  478,  496-514 
Talbot,  A.  N.,  296,  375,  473 
Tangential  stress,  450 
Temperature,  67,  251 
Tempering,  62,  64 
Tensile  tests,  476 
Tension,  2,  9,  36,  69,  407,  496 

and  flexure,  259,  262 

and  shear,  263 

and  torsion,  268 

centrifugal,  164 

eccentric,  73,  262 
Testing  laboratories,  41,  358,  473 
Testing  machines,  40,  433,  470 


INDEX 


523 


Testing,  rules  for,  474-481 
Test  specimens,  15,  436,  474 
Tests,  brick,  49 

cast  iron,  56 
cement,  53,  475 
cold  bend,  58,  239 
columns,  196,  213 
compression,  12,  45,  475,  478 
fatigue,  352 

flexural,  47,  56,  131,  472,  479 
impact,  341,  346,  481 
steel,  63,  474,  482 
stone,  51,  475 
tension,  4,  9,  36,  474,  476 
timber,  47,  187,  438 
torsion,  226,  245,  480 
wrought  iron,  58 
Tetmajer,  L.  von,  208,  224 
Theorem  of  three  moments,  173-183 
Thick  hollow  cylinders,  383,  389,  393 

spheres,  417 
Thin  pipes,  76 
Thurston,  R.  H.,  67,  318 
T'rnber,  5,  10,  13,  14,  24,  38,  46,  475 
beams,  129 

factors  of  safety,  17,  47 
flexural  strength,  131 
resilience,  306 
weight,  42,  46 
Time  of  vibration,  337 
Tool  steel,  65 
Torsion,  225-250,  480 
combined,  152,  154 
formula  for,  228 
in  springs,  444 
non-circular  sections,  245 
phenomena  of,  225 
resilience  of,  310 
rupture  by,  235 
Transmission  of  power,  230 
Transverse  impact,  327,  334 
Trap  rock,  50,  51 
Tredgold,  T.,  40,  347 
Triangular  beams,  104,  106,  166 
Trigonometric  functions,  510 
True  deformations,  360,  370 
stresses,  274,  359-384 
Tubes,  78,  382 


Turner,  C.  A.  P.,  354 
Twisting  moment,  227 

Ultimate  strength,  6,  10,  13,  497 

compression,  12,  44 

deformation,  30 

shear,  14 

tension,  4,  36 

Uniform  load,  88.  \ig,  150 
Uniform  strength,  71,  144 

bars,  71 

beams,  143,  146 
Unit-deformation,  9,  23,  30 
Unit-stress,  i,  12,  23 

repeated,  353 

working,  17 

Unsymmetric  beams,  125,  427 
loads,  428 

Velocity  of  live  load,  350 

stress,  472 

Vertical  shear,  90,  116,  119 
deflection  due  to,  316 
stresses  caused  by,  93,  123 
work  of,  39,  310 
Vibrations  after  impact,  338 

of  a  beam,  344 
Volume,  change  of,  33,  450 

resilience  of,  467 
Volumetric  modulus,  465 

Wagon  springs,  443 
Water,  466,  489 
Water  pipes,  75,  77 
pressure,  76 
Wave  propagation,  489 
Weights  of  bars,  42,  88,  494,  505 
materials,  42-68,  496 
Weyrauch's  formula,  356 
Wheel,  revolving,  422 
Wire,  63,  66 
Wohler's  laws,  353 
Wood,  De  V.,  78,  166,  489 
Work,  least,  320 
Work  of  flexure,  304,  308,  312 

rupture,  36 

shearing,  39 

tension,  35,  307 


524 


INDEX 


Work  of  torsion,  310 

vertical  bar,  70 
vertical  shear,  317 
Working  unit-stresses,  17,  482 
Wrought  iron,  5,  10,  13,  14,  17,  24,  57, 

60,  252,  482 
factors  of  safety,  17 
flexural  strength,  132 
resilience,  307 
shear,  38 
tension,  10,  29,  59 


Wrought  iron,  weight  of,  42,  505 
Wrought-iron  bars,  42,  58,  505 

pipes,  76 

plates,  58 

Yield  point,  27,  29,  63,  443,  479,  482 
Young,  T.,  40,  346,  433 
Young's  modulus,  24 

Z  bars,  104,  427,  430,  503 
Zimmerman,  H.,  351 


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